In the previous section, we saw some solved examples on trigonometric equations. In this section, we will see a few more solved examples.
The list of trigonometric identities can be seen here.
Solved example 3.69
Find the principal solutions and general solution of the equation: sin 2x - sin 4x + sin 6x = 0
Solution:
1. Given that: sin 2x - sin 4x + sin 6x = 0
• Note that, (2+6)/2 = 4. So we will add sin 2x and sin 6x
2. Using identity 20.c, we get:
$\begin{eqnarray}
&{}& \sin 2x + \sin 6x - \sin 4x = 0 \nonumber \\
&\Rightarrow& 2 \sin \left(\frac{2x+6x}{2} \right) \cos \left(\frac{2x-6x}{2} \right) -\sin 4x = 0 \nonumber \\
&\Rightarrow& 2 \sin 4x \cos (-2x) -\sin 4x = 0 \nonumber \\
&\Rightarrow& \sin 4x [2 \cos (-2x) -1] = 0 \nonumber \\
&\Rightarrow& \sin 4x(2 \cos 2x -1) = 0 \nonumber \
\end{eqnarray}$
• We can write:
sin 4x = 0 or (2cos 2x - 1) = 0
3. We will solve this in two parts A and B.
♦ In Part A, we will solve sin 4x = 0
♦ In Part B, we will solve (2cos 2x - 1) = 0
Part A: sin 4x = 0
• First we will find the principal solutions.
1. Given that: sin 4x = 0
2. We know that 0 is a principal solution.
• That means, we can put '0' 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 = 0 is a principal solution.
• When the input x is 0, the left side becomes sin 0
• This gives us: sin 4x = sin 0 = 0
(ii) We want another angle such that, it's sine is also 0
• Such an angle can be calculated using identity 9.d: sin (π-x) = sin x
• We get:
$\sin 0 = \sin (\pi - 0) = \sin \pi$
(iii) Using the results in (i) and (ii), we get:
$\sin 4x = \sin 0 = 0=\sin \pi$
(iv) Picking the first and last items in (iii), we get:
$\sin 4x=\sin \pi$
⇒ $4x=\pi$
⇒ $x=\frac{\pi}{4}$
• Thus we get another value for x
• $0 \leq \frac{\pi}{4} < 2\pi$. So $\frac{\pi}{4}$ (45o) is a principal solution.
• Now we will write the general solution:
1. Given that: sin 4x = 0
• We have to convert this equation into the form: sin x = sin y
• Just now, we saw that: sin 4x = sin 0
2. This is of the form sin x = sin y
♦ In the place of 'x', we have 4x
♦ In the place of 'y', we have 0
• Now we can apply theorem 1:
sin x = sin y implies x = n𝞹 + (-1)ny, where n ∈ Z
• We get: $4x=n\pi + (-1)^n \times 0$, where n ∈ Z
⇒ $x=\frac{n\pi}{4}$, 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 4x = 10
• Table 3.10 below shows some of the solutions:
 |
| Table 3.10 |
• Let us see a sample calculation for the above table:
♦ When n = -3,
♦ x = $\frac{-3 \times 180}{4}=-135$
Part B: (2cos 2x - 1) = 0
1. This can be rearranged as: 2cos 2x = 1
⇒ cos 2x = $\frac{1}{2}$
2. We know that $\frac{\pi}{6}$ is a principal solution.
• That means, we can put '$\frac{\pi}{6}$' in place of x.
3. We will now find another principal solution. It can be done in 4 steps:
(i) We have seen that, $x=\frac{\pi}{6}$ is a principal solution.
• When the input x is $\frac{\pi}{6}$, the left side becomes $\cos \frac{\pi}{3}$
• This gives us: cos 2x = $\cos \frac{\pi}{3}=\frac{1}{2}$
(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:
$\cos 2x = \cos \frac{\pi}{3} = \frac{1}{2} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(iv) Picking the first and last items in (iii), we get:
$\cos 2x=\cos \frac{5\pi}{3}$
⇒ $2x=\frac{5\pi}{3}$
⇒ $x=\frac{5\pi}{6}$
• Thus we get another value for x
• $0 \leq \frac{5\pi}{6} < 2\pi$. So $\frac{5\pi}{6}$ (150o) is a principal solution.
• Now we will write the general solution:
1. Given that: cos 2x = $\frac{1}{2}$
• We have to convert this equation into the form: cos x = cos y
• Just now, we saw that: cos 2x = cos $\frac{\pi}{3}$
2. This is of the form cos x = cos y
♦ In the place of 'x', we have 2x
♦ 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: $2x=2n\pi ± \frac{\pi}{3}$, where n ∈ Z
⇒ $x=n\pi ± \frac{\pi}{6}$, 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 cos 2x = $\frac{1}{2}$
• Tables 3.11 below shows some of the solutions:
 |
| Table 3.11 |
• Let us see a sample calculation for the -ve table above:
♦ When n = -3,
♦ x = (-3 × 180) - 30
♦ = -540 - 30
♦ = -570
4. We have in total, three tables in this problem. We can choose any one x value from those tables. If we input that x, in the LHS of the given equation sin 2x - sin 4x + sin 6x = 0. It will reduce to zero.
An example:
• Let us input x = -135. We get:
• LHS = sin (2 × -135) - sin (4 × -135) + sin (6 × -135)
= sin (-270) - sin (-540) + sin (-810)
= -sin 270 - -sin 540 + -sin 810
= - -1 + 0 - 1
= +1 + 0 - 1
= 0 = RHS
Solved example 3.70
Find the principal solutions and general solution of the equation: 2cos2 x+ 3 sin x = 0
Solution:
1. The given equation can be rearranged as follows:
2(1 - sin2x) + 3 sin x = 0
⇒ 2 - 2sin2x + 3 sin x = 0
⇒ 2sin2x - 3 sin x - 2 = 0
2. Let us put a variable 'u' in place of sin x.
• We will get: 2u2 - 3u - 2 = 0
• This is a quadratic equation in u. Solving it, we get:
u = $-\frac{1}{2}$ or u = 2
3. So we can write: sin x = $-\frac{1}{2}$ or sin x = 2
• But sin x cannot be '2' because, the maximum possible value of sin x is '1'
• So we need to consider only sin x = $-\frac{1}{2}$
4. We know that $\sin \frac{\pi}{6}=\frac{1}{2}$
• We have identity 9.f: sin (𝞹+x) = -sin x
• Using this identity, we can write: $\sin \left(\pi + \frac{\pi}{6} \right)=-\sin \frac{\pi}{6}$
⇒ $\sin \frac{7\pi}{6} =-\sin \frac{\pi}{6}$
• But $\sin \frac{\pi}{6}\;\text{is}\;\frac{1}{2}$
• So we get: $\sin \frac{7\pi}{6} =-\frac{1}{2}$
⇒ $\sin \frac{7\pi}{6} =-\frac{1}{2}=\sin x$
⇒ $x= \frac{7\pi}{6}$
• $0 \leq \frac{7\pi}{6} < 2\pi$. So $\frac{7\pi}{6}$ (210o) is a principal solution.
5. We will now find the other principal solution. It can be done in 6 steps:
(i) We have seen that, x=$\frac{7\pi}{6}$ is a principal solution.
• When the input x is $\frac{7\pi}{6}$, the left side becomes sin $\frac{7\pi}{6}$
• This gives us: sin x = sin $\frac{7\pi}{6} = -\frac{1}{2}$
(ii) We want another angle such that, it's sine is also $-\frac{1}{2}$
• Such an angle can be calculated using identities:
♦ 9.d: sin (π-x) = sin x
♦ 9.h: sin (2π - x) = - sin x
(iii) Using 9.d, we get: $\sin \frac{7\pi}{6}=\sin \left(\pi - \frac{7\pi}{6} \right) = \sin \frac{-1\pi}{6}$
(iv) From (iii), we get: $\sin \frac{7\pi}{6} = \sin \frac{-1\pi}{6}$
• But $\frac{-1\pi}{6}$ is a -ve angle. We want an angle which lies between 0 and 2π.
• So we apply identity 9.h. We get:
$\sin \left(2\pi - \frac{\pi}{6} \right) = -\sin \frac{\pi}{6}$
⇒ $\sin \frac{11\pi}{6} = -\sin \frac{\pi}{6}$
(v) Using (iii) and (iv), we can write:
$\sin \frac{7\pi}{6}=-\sin \frac{\pi}{6}=\sin \frac{11\pi}{6}$
(vi) Thus we get:
$\sin \frac{7\pi}{6}=\sin \frac{11\pi}{6}=\sin x$
⇒ $x=\frac{11\pi}{6}$
• $0 \leq \frac{11\pi}{6} < 2\pi$. So $\frac{11\pi}{6}$ (330o) is the other principal solution.
• Now we will write the general solution:
1. Given that: sin x = $-\frac{1}{2}$
• We have to convert this equation into the form: sin x = sin y
• Just now, we saw that: sin x = sin $\frac{7\pi}{6}$
2. This is of the form sin x = sin y
♦ In the place of 'x', we have x
♦ In the place of 'y', we have $\frac{7\pi}{6}$
• Now we can apply theorem 1:
sin x = sin y implies x = n𝞹 + (-1)ny, where n ∈ Z
• We get: $x=n\pi + (-1)^n \times \frac{7\pi}{6}$, where n ∈ Z
• 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 = $-\frac{1}{2}$
• Table 3.12 below shows some of the solutions:
 |
| Table 3.12 |
• The principal solutions are shown in red color.
• Let us see a sample calculation for the table above:
♦ When n = -3,
♦ x = (-3 × 180) + (-1)-3 × 210
♦ = -540 - 210
♦ = -750
3. We can choose any one
x value from the above table. If we input that x, in the LHS of the given
equation 2cos2 x+ 3 sin x = 0. It will reduce to zero.
An example:
• Let us input x = -150. We get:
• LHS = 2 cos2(-150) + 3 sin (-150)
= $2 \times \left(\frac{\sqrt 3}{2} \right)^2 + 3 \times -\frac{1}{2}$
= $2 \times \frac{3}{4} + 3 \times -\frac{1}{2}$
= $\frac{3}{2} - \frac{3}{2}$
= 0 = RHS
Solved example 3.71
Find the principal solutions and general solution of the equation: sin2 x - cos x = $\frac{1}{4}$
Solution:
1. The given equation can be rearranged as follows:
1 - cos2x - cos x = $\frac{1}{4}$
⇒ cos2x + cos x - $\frac{3}{4}$ = 0
2. Let us put a variable 'u' in place of cos x.
• We will get: u2 + u - $\frac{3}{4}$ = 0
• This is a quadratic equation in u. Solving it, we get:
u = $\frac{1}{2}$ or u = $-\frac{3}{2}$
3. So we can write: cos x = $\frac{1}{2}$ or cos x = $-\frac{3}{2}$
• But cos x cannot be '$-\frac{3}{2}$' because, the least possible value of cos x is '-1'
• So we need to consider only cos x = $\frac{1}{2}$
4. We know that $\cos \frac{\pi}{3}=\frac{1}{2}$
• So we get:
$\cos \frac{\pi}{3} =\frac{1}{2}=\cos x$
⇒ $x= \frac{\pi}{3}$
• $0 \leq \frac{\pi}{3} < 2\pi$. So $\frac{\pi}{3}$ (60o) is a principal solution.
5. 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.
If we input x = $\frac{\pi}{3}$, the left side becomes: cos $\frac{\pi}{3}$
• This gives us: cos 2x = $\cos \frac{\pi}{3}=\frac{1}{2}$
(ii) We want another angle such that, it's cosine is also $\frac{1}{2}$
• Such an angle can be calculated using identitiy 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:
$\cos x = \cos \frac{\pi}{3} = \frac{1}{2} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \frac{5\pi}{3}$
(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 $\frac{5\pi}{3}$ (300o) is a 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
• Just now, we saw that: cos x = cos $\frac{\pi}{3}$
2. This is of the form cos x = cos y
♦ In the place of 'x', we have x
♦ 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: $x=2n\pi ± \frac{\pi}{3}$, where n ∈ Z
• 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 3.13 below shows some of the solutions:
 |
| Table 3.13 |
• The principal solutions are shown in red color.
• Let us see a sample calculation for the -ve table above:
♦ When n = -3,
♦ x = (2 × -3 × 180) - 60
♦ = -1080 - 60
♦ = -1140
3. We can choose any one x value from the above tables. If we input that x, in the LHS of the given equation sin2 x - cos x = $\frac{1}{4}$, It will reduce to $\frac{1}{4}$.
An example:
• Let us input x = -300. We get:
• LHS = sin2 (-300) - cos (-300)
= [sin (-300) × sin (-300)] - cos (-300)
= [-sin 300 × -sin 300] - cos 300
= [-sin 300 × -sin 300] - cos 300
= $\left[\frac{\sqrt 3}{2} \times \frac{\sqrt 3}{2} \right]-\frac{1}{2}$
= $\left[\frac{3}{4}\right]-\frac{1}{2}$
= $\frac{1}{4}$ = RHS
Link to some more solved examples is given below:
In
the next
section, we will see some miscellaneous examples.
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