Chapter 3.17 - Solved Examples on Trigonomeric Equations

In the previous section, we completed a discussion on the three theorems. In this section, we will see how they can be used to solve trigonometric equations.

• The list of trigonometric identities can be seen here.

Solved example 3.63
Find the solution of the trigonometric equation: sin x = $\frac{1}{2}$
Solution:
1. First we have to convert this equation into the form: sin x = sin y
• So we must find a 'y' such that sin y = $\frac{1}{2}$
• We have: $\sin \frac{\pi}{6} = \frac{1}{2}$
   ♦ So we can write $\sin \frac{\pi}{6}$ in the place of $\frac{1}{2}$
• Thus the given equation becomes:
$\sin x=\sin \frac{\pi}{6}$
2. This is of the form sin x = sin y
   ♦ In the place of 'y', we have $\frac{\pi}{6}$
• Now we can apply theorem 1:
sin x = sin y implies x = nπ + (-1)n y, where n ∈ Z
• We get: $x=n\pi+(-1)^n \frac{\pi}{6}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation sin x = $\frac{1}{2}$
Let us see some random examples:
(i) Let us put n = 5
• We get:
$\begin{eqnarray} &{}& x=5\pi+(-1)^{5} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=5\pi+(-1)^{5} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=5\pi+(-1) \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=5\pi - \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=\frac{29\pi}{6} \nonumber \ \end{eqnarray}$
• ${\frac{29\pi}{6}}^c \; \text{is}\; 870^o$
The reader may verify that, sin 870 is $\frac{1}{2}$
(ii) Let us put n = -2
• We get:
$\begin{eqnarray} &{}& x=-2\pi+(-1)^{-2} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=-2\pi+\frac{1}{(-1)^2} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=-2\pi+\frac{1}{1} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=-2\pi+\frac{\pi}{6} \nonumber \\ &\Rightarrow& x=-\frac{11\pi}{6} \nonumber \ \end{eqnarray}$
• ${-\frac{11\pi}{6}}^c \; \text{is}\; -330^o$
The reader may verify that, sin (-330) is $\frac{1}{2}$
4. In this way, infinite number of x values are possible. All 'values of x' thus obtained will satisfy the equation sin x = $\frac{1}{2}$
• Often in scientific and engineering problems, we will want only those 'values of x' which lie in between 0 and 2π.
• For that, we put suitable values of n
5. Let us put suitable values of n in our present case.
We have: $x=n\pi+(-1)^n \frac{\pi}{6}$
(i) Let us put n = 0
• We get:
$\begin{eqnarray} &{}& x=0 \times \pi+(-1)^{0} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=0+1 \times \frac{\pi}{6} \nonumber \\ &{}& \text{(Any number raised to zero is 1)} \nonumber \\ &\Rightarrow& x=\frac{\pi}{6} \nonumber \\ &{}& 0\leq\frac{\pi}{6}<2\pi \nonumber \ \end{eqnarray}$
• So $\frac{\pi}{6}$ (30^o) is acceptable.
(ii) Let us put n = 1
• We get:
$\begin{eqnarray} &{}& x=1 \times \pi+(-1)^{1} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=\pi -1 \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=\frac{5\pi}{6} \nonumber \\ &{}& 0\leq\frac{5\pi}{6}<2\pi\nonumber \ \end{eqnarray}$
• So $\frac{5\pi}{6}$ (150^o) is acceptable.
(iii) Let us put n = 2
• We get:
$\begin{eqnarray} &{}& x=2 \times \pi+(-1)^{2} \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=2\pi +1 \times \frac{\pi}{6} \nonumber \\ &\Rightarrow& x=\frac{13\pi}{6} \nonumber \\ &{}& \frac{13\pi}{6}>2\pi\nonumber \ \end{eqnarray}$
• So $\frac{13\pi}{6}$ (390^o) is not acceptable.
6. Thus the values that lie between 0 and 2π are $\frac{\pi}{6} \; \text{and}\; \frac{5\pi}{6}$
◼ Solutions (values of x) that lie between 0 and 2π are called principal solutions.
• So in our present case, the principal solutions are $\frac{\pi}{6} \; \text{and}\; \frac{5\pi}{6}$
7. The table 3.4 below shows the values of x obtained for various values of n.

Table 3.4

• We see that:
    ♦ When n becomes more and more -ve, x also becomes more and more -ve.
    ♦ When n becomes more and more +ve, x also becomes more and more +ve.
    ♦ The principal solutions are obtained when n = 0 and n= 1
        ✰ They are shown in red color.
8. An easier method to find principal solutions:
• Once we find one of the principal solutions, the other can be calculated using a suitable identity. This can be demonstrated in 4 steps for our present problem:
(i) We have seen that, $x=\frac{\pi}{6}$ is a principal solution.
• When the input x is $\frac{\pi}{6}$, the left side of the given equation becomes: $\sin \frac{\pi}{6}$
• This gives us: $\sin x = \sin \frac{\pi}{6}=\frac{1}{2}$
(ii) We want another angle such that, it's sine is also $\frac{1}{2}$
• Such an angle can be calculated using identity 9.d: sin (𝞹-x) = sin x
• We get:
$\sin \frac{\pi}{6} = \sin \left(\pi - \frac{\pi}{6}\right) = \sin \frac{5\pi}{6}$
(iii) Using the results in (i) and (ii), we get:
$\sin x = \sin \frac{\pi}{6}=\frac{1}{2}=\sin \frac{5\pi}{6}$
(iv) Picking the first and last items in (iii), we get:
$\sin x=\sin \frac{5\pi}{6}$
⇒ $x=\frac{5\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150^o) is the other principal solution.

Solved example 3.64
Find the principal solutions and general solution of the equation: $\sin x=\frac{\sqrt 3}{2}$
Solution:
• First we will find the principal solutions.
1. Given that $\sin x=\frac{\sqrt 3}{2}$
2. We know that $\frac{\pi}{3}$ (same as 60^o) is a principal solution.
• That means, we can put $\frac{\pi}{3}$ in place of x.
3. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{3}$ is a principal solution.
• When the input x is $\frac{\pi}{3}$, the left side becomes $\sin \frac{\pi}{3}$
• This gives us: $\sin x = \sin \frac{\pi}{3}=\frac{\sqrt 3}{2}$
(ii) We want another angle such that, it's sine is also $\frac{\sqrt 3}{2}$
• Such an angle can be calculated using identity 9.d: sin (𝞹-x) = sin x
• We get:
$\sin \frac{\pi}{3} = \sin \left(\pi - \frac{\pi}{3}\right) = \sin \frac{2\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\sin x = \sin \frac{\pi}{3}=\frac{\sqrt 3}{2}=\sin \frac{2\pi}{3}$
(iv) Picking the first and last items in (iii), we get:
$\sin x=\sin \frac{2\pi}{3}$
⇒ $x=\frac{2\pi}{3}$
• Thus we get another value for x
• $0 \leq \frac{2\pi}{3} < 2\pi$. So $\frac{2\pi}{3}$ (120^o) is the other principal solution.

• Now we will write the general solution:
1. Given that: $\sin x=\frac{\sqrt 3}{2}$
• We have to convert this equation into the form: sin x = sin y
• So we must find a 'y' such that sin y = $\frac{\sqrt 3}{2}$
• We have: $\sin \frac{\pi}{3} = \frac{\sqrt 3}{2}$
   ♦ So we can write $\sin \frac{\pi}{3}$ in the place of $\frac{\sqrt 3}{2}$
• Thus the given equation becomes:
$\sin x=\sin \frac{\pi}{3}$
2. This is of the form sin x = sin y
   ♦ In the place of 'y', we have $\frac{\pi}{3}$
• Now we can apply theorem 1:
sin x = sin y implies x = nπ + (-1)n y, where n ∈ Z
• We get: $x=n\pi+(-1)^n \frac{\pi}{3}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation sin x = $\frac{\sqrt 3}{2}$
• Table 3.5 below shows some of the solutions:

Table 3.5

• The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 180) + (-1)^-3 × 60
          = (-3 × 180) + (-1)^-3 × 60
          = -540 + (-1) × 60
          = -540 + - 60
          = -600

Solved example 3.65
Find the principal solutions and general solution of the equation: $\tan x=-\frac{1}{\sqrt 3}$
Solution:
• First we will find the principal solutions.
1. Given that $\tan x=-\frac{1}{\sqrt 3}$
2. We know that $\tan \frac{\pi}{6}=\frac{1}{\sqrt 3}$
• Using identities 9.d and 9.c, we have: tan (π-x) = -tan x 
• So we can write: $\tan \left(\pi -\frac{\pi}{6}  \right)=-\tan \frac{\pi}{6}$
⇒ $\tan \frac{5\pi}{6}=-\tan \frac{\pi}{6}$
3. But $\tan \frac{\pi}{6}=\frac{1}{\sqrt 3}$
• So the result in (2) becomes: $\tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}$
• We can write: $\tan x = \tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}$
• Thus we get: $x=\frac{5\pi}{6}$
• $0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150^o) is a principal solution.
4. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{5\pi}{6}$ is a principal solution.
• When the input x is $\frac{5\pi}{6}$, the left side becomes $\tan \frac{5\pi}{6}$
• This gives us: $\tan x = \tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}$
(ii) We want another angle such that, it's tangent is also $-\frac{1}{\sqrt 3}$
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
$\tan \frac{5\pi}{6} = \tan \left(\pi + \frac{5\pi}{6}\right) = \tan \frac{11\pi}{6}$
(iii) Using the results in (i) and (ii), we get:
$\tan x = \tan \frac{5\pi}{6}=-\frac{1}{\sqrt 3}=\tan \frac{11\pi}{6}$
(iv) Picking the first and last items in (iii), we get:
$\tan x=\tan \frac{11\pi}{6}$
⇒ $x=\frac{11\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{11\pi}{6} < 2\pi$. So $\frac{11\pi}{6}$ (330^o) is a principal solution.
5. So the two principal solutions are:
$\frac{5\pi}{6}\;\; \text{and}\;\;\frac{11\pi}{6}$

• Now we will write the general solution:
1. Given that: $\tan x=-\frac{1}{\sqrt 3}$
• We have to convert this equation into the form: tan x = tan y
• So we must find a 'y' such that tan y = $-\frac{1}{\sqrt 3}$
• Just above, we saw that: $\tan \frac{5\pi}{6} =-\frac{1}{\sqrt 3}$
   ♦ So we can write $\tan \frac{5\pi}{6}$ in the place of $-\frac{1}{\sqrt 3}$
• Thus the given equation becomes:
$\tan x=\tan \frac{5\pi}{6}$
2. This is of the form tan x = tan y
   ♦ In the place of 'y', we have $\frac{5\pi}{6}$
• Now we can apply theorem 3:
tan x = tan y implies x = nπ + y, where n ∈ Z
• We get: $x=n\pi+ \frac{5\pi}{6}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation $\tan x=-\frac{1}{\sqrt 3}$
• Table 3.6 below shows some of the solutions:

Table 3.6

 • The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 180) + 150
          = -540 + 150
          = -540 + - 60
          = -390

Solved example 3.66
Find the principal solutions and general solution of the equation: $\cos x=\frac{1}{2}$
Solution:
• First we will find the principal solutions.
1. Given that $\cos x=\frac{1}{2}$
2. We know that $\frac{\pi}{3}$ (same as 60^o) is a principal solution.
• That means, we can put $\frac{\pi}{3}$ in place of x.
3. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{3}$ is a principal solution.
• When the input x is $\frac{\pi}{3}$, the left side becomes $\cos \frac{\pi}{3}$
• This gives us: $\cos x = \cos \frac{\pi}{3}=\frac{1}{2}$
(ii) We want another angle such that, it's cosine is also $\frac{1}{2}$
• Such an angle can be calculated using identity 9.g: cos (2𝞹-x) = cos x
• We get:
$\cos \frac{\pi}{3} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\cos x = \cos \frac{\pi}{3}=\frac{1}{2}=\cos \frac{5\pi}{3}$
(iv) Picking the first and last items in (iii), we get:
$\cos x=\cos \frac{5\pi}{3}$
⇒ $x=\frac{5\pi}{3}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{3} < 2\pi$. So $\frac{5\pi}{3}$ (300^o) is the other principal solution.

• Now we will write the general solution:
1. Given that: $\cos x=\frac{1}{2}$
• We have to convert this equation into the form: cos x = cos y
• So we must find a 'y' such that cos y = $\frac{1}{2}$
• Just now, we saw that: $\cos \frac{\pi}{3} = \frac{1}{2}$
   ♦ So we can write $\cos \frac{\pi}{3}$ in the place of $\frac{1}{2}$
• Thus the given equation becomes:
$\cos x=cos \frac{\pi}{3}$
2. This is of the form cos x = cos y
   ♦ In the place of 'y', we have $\frac{\pi}{3}$
• Now we can apply theorem 2:
cos x = cos y implies x = 2nπ ± y, where n ∈ Z
• We get: $x=2n\pi \pm \frac{\pi}{3}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation cos x = $\frac{1}{2}$
• Tables 3.7 below shows some of the solutions:

Table 3.7

• The principal solutions are shown in red color.
• Let us see a sample calculation for the first table:
   ♦ When n = -3,
   ♦ x = (2 × -3 × 180) + 60
          = -1080 + 60
          = -1020

Solved example 3.67
Find the principal solutions and general solution of the equation: tan 2x = 1
Solution:
• First we will find the principal solutions.
1. Given that: tan 2x = 1
2. We know that $\frac{\pi}{8}$ (same as 22.5^o) is a principal solution.
• That means, we can put $\frac{\pi}{8}$ in place of x.
3. We will now find another principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{8}$ is a principal solution.
• When the input x is $\frac{\pi}{8}$, the left side becomes $\tan \frac{\pi}{4}$
• This gives us: $\tan 2x = \tan \frac{\pi}{4}=1$
(ii) We want another angle such that, it's tangent is also 1
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
$\tan \frac{\pi}{4} = \tan \left(\pi + \frac{\pi}{4}\right) = \tan \frac{5\pi}{4}$
(iii) Using the results in (i) and (ii), we get:
$\tan 2x = \tan \frac{\pi}{4}=1=\tan \frac{5\pi}{4}$
(iv) Picking the first and last items in (iii), we get:
$\tan 2x=\tan \frac{5\pi}{4}$
⇒ $2x=\frac{5\pi}{4}$
⇒ $x=\frac{5\pi}{8}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{8} < 2\pi$. So $\frac{5\pi}{8}$ (112.5^o) is a principal solution.

• Now we will write the general solution:
1. Given that: tan 2x = 1
• We have to convert this equation into the form: tan x = tan y
• Just now, we saw that: tan 2x = tan $\frac{\pi}{4}$
2. This is of the form tan x = tan y
   ♦ In the place of 'x', we have 2x
   ♦ In the place of 'y', we have $\frac{\pi}{4}$
• Now we can apply theorem 3:
tan x = tan y implies x = nπ + y, where n ∈ Z
• We get: $2x=n\pi +\frac{\pi}{4}$, where n ∈ Z
⇒ $x=\frac{n\pi}{2} +\frac{\pi}{8}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation tan 2x = 1
• Table 3.8 below shows some of the solutions:

Table 3.8

• The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 90) + 22.5
          = -270 + 22.5
          = -247.5

Solved example 3.68
Find the principal solutions and general solution of the equation: $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
Solution:
• First we will find the principal solutions.
1. Given that: $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
• We will write the right side in terms of tan. It can be written as follows:
$\begin{eqnarray} &{}& -\cot\left(x+\frac{\pi}{3}\right) \nonumber \\ &=& \frac{-\cos \left(x+\frac{\pi}{3}\right)}{\sin \left(x+\frac{\pi}{3}\right)} \nonumber \\ &=& \frac{-\;-\;\sin \left(\frac{\pi}{2}+x+\frac{\pi}{3}\right)\;\; \text{(Using identity 9.a)}}{\cos \left( \frac{\pi}{2}+x+\frac{\pi}{3}\right)\;\; \text{(Using identity 9.b)}} \nonumber \\ &=& \frac{\sin \left(x+\frac{5\pi}{6}\right)}{\cos \left(x+\frac{5\pi}{6}\right)} \nonumber \\ &=& \tan \left(x+\frac{5\pi}{6}\right) \nonumber \ \end{eqnarray}$
2. So the given equation becomes: $\tan 2x=\tan \left(x+\frac{5\pi}{6}\right)$
Thus we get: $2x=x+\frac{5\pi}{6}$
⇒ $x=\frac{5\pi}{6}$
$0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150^o) is a principal solution
3. We will now find the other principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{5\pi}{6}$ is a principal solution.
• When the input x is $\frac{5\pi}{6}$, the left side becomes $\tan \frac{5\pi}{3}$
• This gives us: $\tan 2x = \tan \frac{5\pi}{3}=\tan \left(x+\frac{5\pi}{6}\right)$
(ii) We want another angle such that, it's tangent remains the same.
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
$\tan \frac{5\pi}{3} = \tan \left(\pi + \frac{5\pi}{3}\right) = \tan \frac{8\pi}{3}$
(iii) Using the results in (i) and (ii), we get:
$\tan 2x = \tan \frac{5\pi}{3}=\tan \frac{8\pi}{3}=\tan \left(x+\frac{5\pi}{6}\right)$
(iv) Picking the third and last items in (iii), we get:
$\tan \frac{8\pi}{3}=\tan \left(x+\frac{5\pi}{6}\right)$
⇒ $x=\frac{8\pi}{3}-\frac{5\pi}{6}$
⇒ $x=\frac{11\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{11\pi}{6} < 2\pi$. So $\frac{11\pi}{6}$ (330^o) is a principal solution.

• Now we will write the general solution:
1. Given that: $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
• We have to convert this equation into the form: tan x = tan y
• Just now, we saw that: $\tan 2x=\tan \left(x+\frac{5\pi}{6}\right)$
2. This is of the form tan x = tan y
   ♦ In the place of 'x', we have 2x
   ♦ In the place of 'y', we have $\left(x+\frac{5\pi}{6}\right)$
• Now we can apply theorem 3:
tan x = tan y implies x = nπ + y, where n ∈ Z
• We get: $2x=n\pi + x+\frac{5\pi}{6}$, where n ∈ Z
⇒ $x=n\pi +\frac{5\pi}{6}$, where n ∈ Z
3. By putting different values for n, we can obtain different values of x. All 'values of x' thus obtained will satisfy the equation $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
• Table 3.9 below shows some of the solutions:

Table 3.9

• The principal solutions are shown in red color.
• Let us see a sample calculation for the above table:
   ♦ When n = -3,
   ♦ x = (-3 × 180) + 150
          = -540 + 150
          = -390
4. We can choose any x value from the above table. That x value will satisfy the given equation $\tan 2x=-\cot\left(x+\frac{\pi}{3}\right)$
An example:
Let us input x = -570
• Then the LHS becomes: tan (2 × -570) = tan (-1140) = -tan 1140
= -tan (3 × 360 + 60) = -tan 60 = -√3
• RHS becomes: -cot (-570+60) = -cot (510) = -cot (360 + 150) = -cot 150
= -cot (90 + 60) [Using identities 9.a and 9.b] = -tan 60 = -√3
• Thus we get: LHS = RHS

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In the next section, we will see a few more solved examples.

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