Solve for $\theta$: $2 \sin \theta = 2 - \cos \theta$ Actually , I'm new to trigonometry .. So i want help in this question


  $$2 \sin \theta = 2 - \cos \theta $$


My attempt ->

$$\begin{align}
2 \sin \theta &amp;= 2 - \cos \theta \\
2 \sin \theta &amp;= 2 - {(1-\sin^{2}\theta)}^{1/2}
\end{align}$$
Beware that substituting $\cos\theta$ with $\sqrt{1-\sin^2\theta}$ is not correct. Beware also that squaring will add spurious solutions.



Here's a method: set $x=\cos\theta$ and $y=\sin\theta$, so you have
$$
\begin{cases}
2y=2-x \\[4px]
x^2+y^2=1
\end{cases}
$$
This basically transforms the problem into finding the intersection of a line with a circle.

Solving the first with respect to $x$ gives $x=2-2y$, so
$$
4-8y+4y^2+y^2=1
$$
becomes
$$
5y^2-8y+3=0
$$
yielding
$$
y=\frac{3}{5}
\qquad\text{or}\qquad
y=1
$$
The first solution gives $x=4/5$ and the second solution gives $x=0$. Finally you have
$$
\theta=\arcsin\frac{3}{5}+2k\pi \quad(k\text{ an integer})
$$
or
$$
\theta=\frac{\pi}{2}+2k\pi \quad(k\text{ an integer})
$$
Another way to solve this problem is to divide both sides by $\cos \theta$.
$$2\sin\theta = 2 - \cos\theta$$
$$2 = 2\sin\theta + \cos\theta$$
$$\frac{2}{\cos\theta} = \frac{2\sin\theta + \cos\theta}{\cos\theta}$$
$$2\sec\theta = 2\tan\theta + 1$$
$$\sec\theta = \tan\theta + \frac{1}{2}$$
$$\sec^2\theta = (\tan\theta + \frac{1}{2})^2$$
$$\sec^2\theta = \tan^2\theta +\tan\theta + \frac{1}{4}$$
$$\sec^2\theta = (\tan^2\theta +1)-1+\tan\theta + \frac{1}{4}$$
$$\sec^2\theta = \sec^2\theta-1+\tan\theta + \frac{1}{4}$$
$$\tan\theta = \frac{3}{4}$$