Question

3(1-sinx) = 2cos2x

Answer 1

what are you looking for in the equation?

Answer 2

\[
3 (1-\sin (x))=2 \cos (2 x) =\\
3-3 \sin (x)=2 \left(\cos ^2(x)-\sin
^2(x)\right)\\
3-3 \sin (x)=2 \left(1-\sin ^2(x)-\sin
^2(x) \right)\\
3-3 \sin (x)=2 \left(1-2 \sin ^2(x)\right)
\]
Can you finish it now?

Answer 3

\[

3-3 \sin (x)=2-4 \sin ^2(x)\\
4 \sin ^2(x)-3 \sin (x)+1=0\\

\]

Answer 4

The last equation have only imaginary roots for sin(x). There are no real solutions