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

1 Answer
Feb 15, 2017

#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#