Question

How do you solve `sin^2(2x)+cos^2x=0`?

Answer 1

For (x, 2pi)
`pi/6; pi/2; (5pi)/6; (7pi)/6; (3pi)/2; (11pi)/6`

Explanation:

Reminder:
`sin^2 (2x) = 4sin^2 x.cos^2 x`
We get:
`sin^2 (2x) + cos^2 x = cos^2 x(1 + 4sin^2 x) = 0`
a. `cos^2 x = 0` --> cos x = 0 --> 2 solutions:
`x = pi/2` and `x = (3pi)/2`
b. `1 + 4sin^2 x = 0`--> `sin^2 x = 1/4` -->
`sin x = +- 1/2`
Use trig table and unit circle.
`sin x = 1/2` --> `x = pi/6` and `x = (5pi)/6`
`sin x = - 1/2` -->`x = (7pi)/6` and `x = (11pi)/6`