Find the solutions of: $\sin x+\cos x=\sin^2 x+0.5\sin{2x}$

Question

Find the solutions of: $\sin x+\cos x=\sin ^2 x+0.5\sin{(2x)}$

How can I find the solutions ?

Answer 1

Your equation, as you wrote it, is not true for all $x \in \mathbb{R}$, nor even for most of them. Consider $x=0$, in which case the left hand side evaluates to $1$ and the right hand side equals $0$.

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We can still find what values of $x$, if any, satisfy your equation.

Recall that

$$\sin (2x) = 2\sin x \cos x$$

Substitute this into your equation, to get

$$\sin x+\cos x=\sin ^2 x+ (0.5 \cdot 2)(\sin x \cos x)$$

$$\sin x+\cos x=\sin ^2 x+ \sin x \cos x $$

$$\sin x+\cos x= (\sin x)(\sin x + \cos x)$$

This equation is true when either

$$\sin x = 1$$

or

$$\sin x + \cos x = 0$$

The first condition occurs when $x=2\pi n + \frac{\pi}{2}$ for some $n \in \mathbb{Z}$. The second condition holds whenever $x= \pi n - \frac{\pi}{4}$ for some $n \in \mathbb{Z}$.