Question

Cos A = Sin 2A solve it

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

Given,

$\large{ \sf{cos \: x = sin2x}}$

Now,

$\large{ \implies \: \sf{2sin \: x.cos \: x - cos \: x = 0}} \\ \\ \large{ \implies \: \sf{2sin \: x - 1 = 0}} \\ \\ \large{ \implies \: \sf{2sin \: x = 1}} \\ \\ \large{ \implies \: \sf{sin \: x = \frac{1}{2} }} \\ \\ \large{ \implies \: \sf{sin \: x = sin \: \frac{ \pi}{6} }} \\ \\ \huge{ \implies \: \boxed{ \boxed{ \tt{x = \frac{\pi}{6} }}}}$

Also,

$\large{ \sf{sin \: x = sin( \frac{6\pi - \pi}{6}) }} \\ \\ \large{ \longrightarrow \: \boxed{ \boxed{\sf{sin \: x = sin \: \frac{5\pi}{6} }}}} \\ \\ \huge{ \rightarrow \: \sf{x = \frac{5\pi}{6} }}$

The principal solutions of given function would be π/6 and 5π/6

General Solution

$\huge{ \boxed{ \boxed{ \tt{x = n \frac{\pi}{6} + ( - 1) {}^{n} }}}}$

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