Question

How do you solve `sin(2x)=-cos(2x)`?

Answer 1

`x = (3pi)/8 + kpi`
`x = (7pi)/8 + kpi`

Explanation:

sin 2x + cos 2x = 0
Use trig identity: `sin a + cos a = sqrt2cos (a - pi/4)`
In this case:
`sin 2x + cos 2x = sqrt2cos (2x - pi/4) = 0`
`cos (2x - pi/4) = 0`
Unit circle gives 2 solutions:
a. `2x - pi/4) = pi/2 + 2kpi` --> `2x = pi/2 + pi/4 = (3pi)/4 + 2kpi` -->
`x = (3pi)/8 + kpi`.
b. `(2x - pi/4) = (3pi)/2 + 2kpi` --> `2x = (3pi)/2 + pi/4 = (7pi)/4 + 2kpi` -->`x = (7pi)/8 + kpi`