Question

Solve . sin^2(2x)+sin^2(3x)+sin^2(4x)+sin^2(5x)=2

Answer 1

\[2(\sin^2(2x)+\sin^2(3x)+\sin^2(4x)+\sin^2(5x)=2)\]
\[1-\cos4x + 1-\cos6x+1-\cos8x+1-\cos10x=4\]
\[\cos4x + \cos 6x + \cos8x+\cos10x=0\]
summation gives
\[\frac{\sin (\frac{4(2x)}{2})\cos(4x+\frac{3(2x)}{2})}{\sin 2x}=0\]

Answer 2

And after summation?

Answer 3

Solve for x

Answer 4

Could you do the cos part ?

Answer 5

cos(7x)=0
\[7x=2n \pi \pm \pi/2\]

Answer 6

Thank you!

Answer 7

Nicely done NSB.