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The number of values of $\displaystyle x \in (-90^{\circ},90^{\circ})$ such that $x\neq 36^{\circ}n$ for $n=0,\pm 1, \pm 2$ and

$\tan x = \cot 5x$ as well as $\sin 2x = \cos 4x$.

Attempt: $\displaystyle \tan x = \cot \left(90^{\circ}-5x\right)$ so $x = n\pi +(90^{\circ}-5x)$ So $\displaystyle x= 36^{\circ}+18^{\circ}$

and $\cos 4x = \cos(90^{\circ}-2x)$ so $4x = 2m\pi\pm 2x$ so $x=60^{\circ}m$ and $x=180^{\circ}m$

now from $x=36^{\circ}n+18^{\circ}$ we have $x=-18^{\circ},18^{\circ}$

and from $x=60^{\circ}m$ we have $x=0,\pm 60^{\circ}$

so we have $6$ solutions but answer is $3$ solution.

could some help me, thanks

DXT
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1 Answers1

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Actually, $$x = n\pi + \frac {\pi}{2}-5x $$ $$\Rightarrow 6x = n\pi + \frac {\pi}{2} $$ $$x = 30^\circ n + 15^\circ $$

Anyways, beside the typo, the question asks for solutions that satisfy both equations and not all solutions, that is, if $x_1$ is a solution, then both $(1) $ and $(2) $ should satisfy. $$\tan x_1 =\cot 5x_1 \tag{1} $$ $$\sin 2x_1 =\cos 4x_1 \tag {2} $$

Check that there are only three such $x'$s that satisfy the condition, $-45^\circ, 15^\circ, 75^\circ $. Hope it helps.