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.
 A: 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, $x = 2n\pi + \dfrac{3 \pi}2$.
A: Hint: Square both sides, and express $\cos^2 x$ in terms of $\sin^2 x$. 
You will get a quadratic equation in $\sin^2 x$. After finding the solutions of this equation, make sure to substitute into the original equation to check whether they are indeed roots of the original equation. The squaring process could produce "spurious" roots. (It happens not to, but checking is cheap.)
A: This may be the hard way, but if you let $q=\tan(x/2)$, then $\sin x=2q/(1+q^2)$, and $\cos x=(1-q^2)/(1+q^2)$. 
A: Let me take a stab at this.  I had to double check a few identities.  First, we use the identities $\cos(\frac\pi 2-x)=\sin x$ and $\sin(\frac\pi 2-x)=\cos x$.
$$1+\cos(\frac\pi 2-x)=2\sin(\frac\pi 2-x)$$
$$\frac{1+\cos(\frac\pi 2-x)}{\sin(\frac\pi 2-x)}=2$$
Next, we use the half-angle identity $\tan\frac\theta 2=\frac{\sin\theta}{1+\cos\theta}$.
$$\cot(\frac\pi 4-\frac x2)=2$$
$$\tan(\frac\pi 4-\frac x2)=\frac12$$
From here, take the inverse tangent and the rest is algebra.
A: We can try to avoid squaring to avoid tests for extraneous root.
$$2\cos x-\sin x=1$$
Putting $r\cos\theta=2,r\sin\theta=1$ where $r>0 $
Squaring and adding we get $r^2=2^2+1^2\implies r=\sqrt5$
On division, $\tan \theta=\frac12$ where $0<\theta<\frac\pi2$ as $\sin\theta,\cos\theta>0$
we get $$\cos\theta\cos x-\sin\theta\sin x=\sin\theta$$(cancelling $r$ in either sides )
or, $\cos(x+\theta)=\cos(\frac\pi2-\theta)$
So, $$x+\theta=2n\pi\pm\left(\frac\pi2-\theta\right)$$
 where $n$ is any integer.
Taking $'-'$ sign, $$x+\theta=2n\pi-\left(\frac\pi2-\theta\right)\implies x=2n\pi-\frac\pi2$$
Taking $'+'$ sign, $$x+\theta=2n\pi+\left(\frac\pi2-\theta\right)\implies x=2n\pi+\frac\pi2-2\theta=2n\pi+\frac\pi2-2\arctan \frac12$$
Now, we know $\cos2y=\frac{1-\tan^2y}{1+\tan^2y},\sin2y=\frac{2\tan y}{1+\tan^2y},\tan2y=\frac{2\tan y}{1-\tan^2y}$
If $\arctan \frac12=y,\tan y=\frac 12$
$$\cos2y=\frac35,\sin2y=\frac45,\tan2y=\frac43$$
So, $$2y=2\arctan \frac12=\arccos\frac35=\arcsin\frac45=\arctan \frac43$$
So, $$x=2n\pi+\frac\pi2-\arccos\frac35=2n\pi+\arcsin\frac35$$ as $\arcsin z+\arccos z=\frac12$ for  $-1\le z\le1$
Similarly, $$x=2n\pi+\arccos\frac45=2n\pi+\operatorname{arccot}\frac43$$
