The-number-of-values-of-x-lying-in-pi-pi-and-satisfying-2-sin-2-cos-2-and-sin-2-2-cos-2-cos-1-0-is-

Question Number 17492 by Tinkutara last updated on 06/Jul/17

The number of values of x lying in

[- π, π] and satisfying 2 \sin^{2} θ = \cos 2 θ

and \sin 2 θ + 2 \cos 2 θ - \cos θ - 1 = 0 is

Answered by ajfour last updated on 08/Jul/17

2 \sin^{2} θ = \cos 2 θ

\Rightarrow 1 - \cos 2 θ = \cos 2 θ

\Rightarrow \cos 2 θ = \frac{1}{2}

\sin 2 θ + 2 \cos 2 θ = 1 + \cos θ

with \cos 2 θ = 1 / 2, it becomes

2 \sin θ \cos θ = \cos θ

\Rightarrow \sin θ = \frac{1}{2}

In [- π, π] \cos 2 θ = \frac{1}{2}, \sin θ = \frac{1}{2}

\Rightarrow θ = \frac{π}{6}; and θ = \frac{5 π}{6} .

Commented by mrW1 last updated on 07/Jul/17

I think θ = \frac{π}{6} and \frac{5 π}{6}

Commented by Tinkutara last updated on 08/Jul/17

Thanks Sir!