Question

Find number of value of $xϵ[0,\pi ]$ satisfying the relation cos 3x + sin 2x - sin 4x = 0

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Answer 1

We have to simplify the given expression to solve it. By looking at the expression, we see that we can

apply transformation formula for sin 2x — sin 4x.

$\Rightarrow$ cos 3x + sin 2x - sin 4x = 0

$\Rightarrow$ cos 3x + 2 sin (-x) cos 3x = 0

$\Rightarrow$ cos 3x (1 - 2 sin x) = 0

$\Rightarrow$ cos 3x = 0 or (1-2 sin x) = 0
$\Rightarrow$ cos 3x = 0 or sin x = $\frac{1}{2}$
$\Rightarrow$ 3x = (2n + 1) $\frac{\pi }{2}$ or $x=n\pi$ + $(-1{)}^n$ $\frac{\pi }{6}$
$\Rightarrow$ x = (2n + 1) $\frac{\pi }{6}$ or $x=n\pi$+ $(-1{)}^n$ $\frac{\pi }{6}$
By using these two expressions we see that the value of x which lie in $[0,\pi ]$ in are $\frac{\pi }{6}$ , $\frac{\pi }{3}$ , $\frac{\pi }{3}$ (we get these
values by substituting n = 1, 2, 3 .... in the expressions we got for x)
So we get three values in $[0,\pi ]$