Question

Find the number of solutions of the equation sin 5x cos 3x = sin 6x cos 2x,
${\text{x}} \in {\text{[0,}}\pi {\text{]}}$.

Answer 1

Hint: Here we need to use the formulas of (2 sin(a) cos(b)) and sin2P using this and considering the values under ${\text{x}} \in {\text{[0,}}\pi {\text{]}}$ you will get the right answer.

Complete step-by-step answer:
So, the given equation is sin 5x cos 3x = sin 6x cos 2x.
Multiplying with 2 both sides we get,
2sin 5x cos 3x = 2sin 6x cos 2x ……….(1)
As we know that 2sinAcosB=sin(A+B)+sin(A-B)
Using the same in (1) we get,
$\Rightarrow$ sin(5x + 3x)+sin(5x – 3x)=sin(6x + 2x)+sin(6x – 2x)
$\Rightarrow$ sin8x+sin2x = sin8x+sin4x
On cancelling sin8x we get,
sin2x = sin4x………….(2)
We know that sin2P = 2 sinP cosP.
So, sin4x = sin2(2x)=2 sin2x cos2x
On putting the value of sin4x in equation (2) we get,
$\Rightarrow$ sin2x = 2 sin2x cos2x
On cancelling sin2x from both the sides we get,
$\Rightarrow$ 2cos2x = 1
Then, $\cos {\text{2x = }}\dfrac{1}{2}$

It is given that
$
  0 \leqslant {\text{x}} \leqslant \pi \\
  {\text{then}} \\
  0 \leqslant 2{\text{x}} \leqslant 2\pi \\
$
And we know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2} = \cos 2{\text{x}}$.
The general solution of cos is ${\text{2n}}\pi \pm {\text{a}}$ where a is the angle.
So, the equation $\cos \dfrac{\pi }{3} = \cos 2{\text{x}}$
Can be written as ${\text{2x = 2n}}\pi \pm \dfrac{\pi }{3}$
Then ${\text{x = n}}\pi \pm \dfrac{\pi }{6}$.
Hence, the answer is ${\text{x = n}}\pi \pm \dfrac{\pi }{6}$.

Note: To solve such questions we need to recall the concept of general solution of trigonometric angles. Using that you can proceed to get the answer correct. You can also check the values by changing the value of n in the general solution. Doing this will solve your problem.