Question

How do you solve sin2x = √2 cosx, 0≤x≤ π?

sin2x = `sqrt(2)` cosx, 0`<=`x`<=` `pi`?

Answer 1

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

Explanation:

Use trig identity : sin 2x = 2sin x.cos x
Substitute in the equation sin 2x by 2sin x.cos x
`2sin x.cos x - sqrt2cos x = 0`
`cos x(2sin x - sqrt2) = 0`
Either one of the 2 factors should be zero:
a. cos x = 0 --> `x = pi/2`, and `x = (3pi)/2`
b. `2sin x - sqrt2 = 0`
`sin x = sqrt2/2`
Trig unit circle gives 2 solutions:
`x = pi/4 and x = (3pi)/4`
Answers for `(0, pi)`
`pi/2; pi/4; (3pi)/4`