Question

If sin 3x + cos 2x = -2, then find the value of x

• A. (2m+1) $\frac{\pi }{2}$
• B. (4m-1) $\frac{\pi }{6}$
• C. (4m+1) $\frac{\pi }{2}$
• D. (2m+1) $\frac{\pi }{6}$

Answer 1

The correct option is C

(4m+1) $\frac{\pi }{2}$

Minimum value of sin 3x and cos 2x is -1. If sin 3x + cos 2x is -2, both sin 3x and cos 2x should be equal to -1.

[This is because if any of them is greater than -1, the other should be less than -1, which will lead to contradiction].
$\Rightarrow$ sin 3x = -l and cos 2x = -1

We want to find the value of x which will satisfy both.

sin 3x = -1 $\Rightarrow$ 3x = (4n - 1) $\frac{\pi }{2}$

$\Rightarrow$ x = (4n - 1) $\frac{\pi }{6}$ ........... (1)

cos 2x = - 1 $\Rightarrow$ 2x = (2m + 1)$\pi$
$\Rightarrow$ x = (2m + 1) $\frac{\pi }{2}$ ...........(2)

Now, we want to find the value of x common to both (2m + 1) $\frac{\pi }{2}$ and (4n - 1) $\frac{\pi }{6}$. That means for some

value of m and n, they should be equal.
$\Rightarrow$ (2m + 1) $\frac{\pi }{2}$= (4n - 1) $\frac{\pi }{6}$
$\Rightarrow$ 6m+ 3 = 4n-1
$\Rightarrow$ n =$\frac{6m+4}{4}$

$\Rightarrow$ n = $\frac{3}{2}$m + 4 ..............(3)

If m = 0,2,4.......... we will have corresponding n for each value of m.

$\Rightarrow$ m = should be of the form 2p.

$\Rightarrow$ x = (2 × 2p +1) $\frac{\pi }{2}$ (from (2) )

= (4p + 1) $\frac{\pi }{2}$