The-solution-of-the-equation-cos-2-2cos-4sin-sin2-where-0-pi-is-

Question Number 17151 by Tinkutara last updated on 01/Jul/17

The solution of the equation

\cos^{2} θ - 2 \cos θ = 4 \sin θ - \sin 2 θ where

θ \in [0, π] is

Answered by ajfour last updated on 01/Jul/17

4 \sin θ - 2 \sin θ \cos θ + 2 \cos θ - \cos^{2} θ = 0

2 \sin θ (2 - \cos θ) + \cos θ (2 - \cos θ) = 0

(2 - \cos θ) (2 \sin θ + \cos θ) = 0

\Rightarrow 2 \sin θ + \cos θ = 0

or \tan θ = - \frac{1}{2}

In [0, π], θ = π - \tan^{- 1} (\frac{1}{2}) .

Commented by Tinkutara last updated on 01/Jul/17

Thanks Sir!