Question 429870
Solution:From the trigonometry we know that: {{{cos2x=(cos(x))^2-(sin(x))^2}}}

 {{{(cos(x))^2=1-(sin(x))^2}}}, and cos(90+x)=-sin(x). Now we write:

{{{-sin(x)-((cos(x))^2-(sin(x))^2)=-sin(x)+(sin(x))^2-(1-(sin(x))^2)=0}}}

{{{2(sin(x))^2-sin(x)-1=0}}}, substitute six=a and rewrite the equation:

{{{2a^2-a-1=0}}}, which is a quadratic equation. Solve this equation:

There are two roots: a=1 and a=-1/2, thus we have: sin(x)=1 and sin(x)=-12

Solve now these two trigonometric equations and find:

x=2n*{{{pi}}}+ {{{pi}}}/2, 

x=2n*{{{pi}}}-{{{pi}}}/6, and x=2n*{{{pi}}}+{{{7pi}}}/6

Done.