Question

How do you find the exact value of `sin^2x-cos^2x=0` in the interval `0<=x<360`?

Answer 1

`pi/4 ; (3pi)/4 ; (5pi)/4 ; (7pi)/4`

Explanation:

Use trig identity: `sin^2 x - cos^2 x = - cos 2x`
Solve the equation:
- cos 2x = 0
Use trig unit circle:
a. cos 2x = 0 --> `2x = pi/2 + 2kpi` --> `x = pi/4 + kpi`
If k = o --> `x = pi/4`
If k = 1 --> `x = pi/4 + pi = (5pi)/4`
b. cos 2x = 0 --> `2x = (3pi)/2 + 2kpi` --> `x = (3pi)/4 + kpi`
If k = o --> `x = (3pi)/4`
If k = 1 --> `x = (3pi)/4 + pi = (7pi)/4`
Answers for `(0, 2pi)`
`pi/4 ; (5pi)/4 ; (3pi)/4 ; (7pi)/4`