Solving a trig equation $1+ \sin (x)=2 \cos(x)$? How would I solve the following equation?
$$
1+ \sin (x)=2 \cos(x)
$$
I am having difficult with it.

We have $1 + \sin(x) = 2 \cos(x)$. Recall that $1 - \sin^2(x) = \cos^2(x)$. Hence, we get that $$(1 + \sin(x))(1- \sin(x)) = \cos^2(x)$$ i.e. 
$$2 \cos(x) (1 - \sin(x)) = \cos^2(x)$$
If $\cos(x) \neq 0$, then we get that $$1 - \sin(x) = \dfrac{\cos(x)}2$$
Hence, have
\begin{align}
1 + \sin(x) & = 2\cos(x)\\
1 - \sin(x) & = \dfrac{\cos(x)}2
\end{align}
This gives us that $2 = \dfrac52 \cos(x) \implies \cos(x) = \dfrac45$. This gives us $\sin(x) = \dfrac35$. Hence, we get one possible solution as $$x = 2n \pi + \theta$$ where $\sin(\theta) = \dfrac35$ with $0 < \theta < \dfrac{\pi}2$.
If $\cos(x) = 0$, then we need $\sin(x) = -1$. Hence, $x = 2n\pi + \dfrac{3 \pi}2$.
