Question

How do I find ALL solutions to the following equation?

`sin^2xcos^2x=(2-sqrt2)/16`

Answer 1

`x = +- 37^@08`
`x = +- 52^@92`

Explanation:

Use trig identity: sin 2x = 2sin x.cos x
`sin^2 2x = 4sin^2 x.cos ^2 x`
`sin^2 x.cos^2 x = (sin^2 2x)/4 = (2 - sqrt2)/16`
`sin^2 2x = (4(2 - sqrt2))/32 = (2 - sqrt2)/8`
`sin 2x = +- sqrt(2 - sqrt2)/(2sqrt2)`
Calculator gives:
`2x = +- 0.77/2.82 = +- 0.273`
Calculator and unit circle give 4 solutions:
a. 2x = 0.273 --> `2x = +- 74^@15` -->`x = +- 37^@08`
b. 2x = - 0.273 --> `2x = +- 105.84` --> `x = +- 52^@92`