In triangle `A B C ,` angle `A` is greater than angle `Bdot` If the measure of angles `Aa n dB` satisfy the equation `3sinx-4sin^3x-k=0,0

Options

A. `(pi)/(3)`
B. `(pi)/(2)`
C. `(2pi)/(3)`
D. `(5pi)/(6)`

Answer

C

Explanation

Given `3 sin x - 4 sin^(3) - 4 sin^(3) x = k, 0 lt k lt 1` which can also be written as `sin 3x = k`
It is given that A and B are solutions of this equation.
Therefore,
`sin 3a = k and sin 3B= k` where `0 lt k lt 1`
`rArr " " 0 lt 3A lt pi and 0 lt 3B lt pi`
Now, `sin 3A = k and sin 3B = k`
`rArr " " sin 3A - sin 3B = 0`
`rArr " " 2cos.(3)/(2) (A+ B) sin .(3)/(2) (A-B) = 0`
`rArr cos 3((A+B)/(2)) = 0 , sin ((A-B)/(2)) =0`
But it is given that `A gt B` and `0 lt 3A lt pi, 0 lt 3B lt pi`
Therfore ,`sin 3((A-B)/(2)) ne 0`
Hence, `cos3((A+B)/(2)) = 0`
`rArr 3((A+B)/(2)) = (pi)/(2)`
`rArr " "A+ B=(pi)/(3)`
`rArr " "C = pi - (A +B) = (2pi)/(3)`