Question

How do you solve `cosx=sin2xsinx`?

Answer 1

`x=pi/4 or 45^@`

Explanation:

recall that `sin2x=2sinxcosx`

`cosx=sin2xsinx`
`cosx=2sinxcosxsinx`
`1=2sin^2x`
`sin^2x=1/2`

`sinx=1/sqrt2`

`x=45^@`

Answer 2

`pi/2; (3pi)/2; pi/4; (3pi)/4; (5pi)/4; (7pi)/4`

Explanation:

cos x = sin 2x.sin x
`cos x = 2sin x.cos x.sin x = 2cos x.sin^2 x`
`cos x - 2cos x.sin^2 x = 0`
Put cos x in common factor:
`cos x(1 - 2sin^2 x) = 0`
Use trig table, and unit circle to solve this trig equation
a. cos x = 0 --> `x = pi/2 and x = (3pi)/2`
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 = (5pi)/4 and x = (7pi)/4`

Answers for `(0, 2pi)`:
`pi/2; (3pi)/2; pi/4; (3pi)/4; (5pi)/5; (7pi)/4.`
For general answers, add `2kpi`