Question

How do you solve `cosx-sinxsin2x=0`?

Answer 1

`0, pi/4, (3pi)/4, (5pi)/4, (7pi)/4, 2pi`

Explanation:

Use trig identity:
sin 2x = 2sin x.cos x.
Substitute this value into the equation:
cos x - sin x (2sin x.cos x) = 0.
Put cos x into common factor:
`cos x(1 - 2sin^2 x) = 0`
Solve it by using trig table and unit circle:

a. cos x = 0 --> x = 0 and `x = 2pi`
b. `(1 - 2sin^2 x) = 0`
`sin^2 x = 1/2`
`sin x = +- 1/sqrt2 = +- sqrt2/2`
c. `sin x = sqrt2/2` --> `x = pi/4 and x = (3pi)/4`
d. `sin x = -sqrt2/2` --> `x = - pi/4 and x = -(3pi)/4`
Arc `(-pi/4)` is co-terminal to arc `(7pi)/4`
Arc `((-3pi)/4)` is co-terminal to arc `((5pi)/4)`
Answers for `(0, 2pi)`
`0, pi/3, (3pi)/4, (5pi)/4, (7pi)/4, 2pi`