Question

How do you find all solutions of the equation `sin^2x cos^2x=2-sqrt 2/16`?

Answer 1

Solve trig equation: `sin^2 x.cos^2 x = 2 - sqrt2/16`

Explanation:

Use trig identity: sin 2x = 2sin x.cos x

`(sin^2 (2x)/4) = (2 - 0.09)/4 = 1.91/4 = 0.48`
On the trig unit circle,

`sin 2x = +- 0.69` . There are 4 answers within period`(0, 2pi)`

1. sin 2x = 0.69 -> 2x = 43.71 --> x = 21.86 deg

2. sin 2x = 0.69 -> 2x = 180 - 43.71 = 136.29 -> x = 68.14

3. sin 2x = - 0.69 -> 2x = 180 + 43.71 = 223.71 -> x = 111.86

4. sin 2x = - 0.69 --> 2x = 360 - 43.71 = 316.29 -> x = 158.16