Solve for x if squareroot 2 sinpi2 x sin2x over the interv

Solve for x, if squareroot 2 sin(pi/2 - x) = sin2x over the interval [0,2pi] Solve for x, if 5sin^2x - cos^2x =cos 2x over the interval [0, 360 degree]

Solution

5.)

sin(pi/2-x)=cos x

sqrt 2 cosx=sin 2x

we know sin2x=2 sinx cosx

so

sqrt 2=2 sin x

sin x=sqrt2 / 2=1/sqrt 2

sin x= sin 45

x=45 degree