Question

The angle `theta` satisfies `tan(2theta)=sintheta`. Show that either `sintheta=0` or `2costheta=2cos^2theta-1` (i did this). Hence find the smallest possible value of `theta`?

Answer 1

Solve `tan 2x = sin x`

Explanation:

`sin 2x/(cos 2x) = sin x`
`sin 2x = sin x.cos 2x`
`2sin x.cos x - sin x(2cos^2 x - 1)`
`sin x(2cos x - 2cos^2 x + 1) = 0`
Either `sin x = 0` or `(-2cos^2 x + 2cos x + 1) = 0`
Call cos x = t, we get:
`-2t^2 + 2t + 1 = 0`
`D = d^2 = 4 + 8 = 12 = 4sqrt3` --> `d = +- 2sqrt3`
`t = -2/4 +- (2sqrt3)/4 = (-1 +- sqrt3)/2`

The smallest possible of arc x has as cos:

`cos x = t = (-1 + sqrt3)/2 = (0.73)/2 = 0.36`
cos x = 0.36 --> x = 68,90
Smallest value of x -->`x = 68^@90`