Question

cos 2x+sin x=1 find all solutions between 0 and 2pi

Answer 1

\begin{align*}
\cos(2x) &= Re(e^{2xi}) \\
&= Re((e^{xi})^2)\\
&= Re(\cos^2(x) - \sin^2(x) + 2i\cos(x)\sin(x))\\
&= \cos^2(x) - \sin^2(x)
\end{align*}
\[ \cos(2x) + \sin(x) = 1\]\[\implies \cos^2(x) - \sin^2(x)+ \sin(x) = 1\]
Using \(\cos^2(x) + \sin^2(x) = 1\)
\[\implies (1-\sin^2(x)) - \sin^2(x) + \sin(x) = 1\]\[\implies \sin(x) - 2\sin^2(x) = 0\]\[\implies(\sin x)(1-2\sin(x)) = 0\]
\[\implies \begin{array}{ccc}\sin x = 0 & \text{OR} & \sin x = \frac{1}{2}\end{array}\]

Answer 2

Cos 2x is 1 -2 sin^2 x gives the quadratic in sin x directly.