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 2θ and sin 2θ + 2 cos 2θ − cos θ − 1 = 0 is](Q17492.png)
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{lying}\:\mathrm{in} \\ $$$$\left[−\pi,\:\pi\right]\:\mathrm{and}\:\mathrm{satisfying}\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:=\:\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\mathrm{and}\:\mathrm{sin}\:\mathrm{2}\theta\:+\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta\:−\:\mathrm{cos}\:\theta\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 08/Jul/17
![2sin^2 θ=cos 2θ ⇒1−cos 2θ=cos 2θ ⇒ cos 2θ=(1/2) sin 2θ+2cos 2θ=1+cos θ with cos 2θ=1/2, it becomes 2sin θcos θ=cos θ ⇒ sin θ=(1/2) In [−π, π] cos 2θ=(1/2), sin θ=(1/2) ⇒ θ= (π/6) ; and θ=((5π)/6) .](Q17537.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2sin}\:^{\mathrm{2}} \theta=\mathrm{cos}\:\mathrm{2}\theta\: \\ $$$$\Rightarrow\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta=\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\Rightarrow\:\:\:\:\:\:\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{sin}\:\mathrm{2}\theta+\mathrm{2cos}\:\mathrm{2}\theta=\mathrm{1}+\mathrm{cos}\:\theta \\ $$$$\:\mathrm{with}\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}/\mathrm{2},\:\:\mathrm{it}\:\mathrm{becomes} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2sin}\:\theta\mathrm{cos}\:\theta=\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{In}\:\:\:\left[−\pi,\:\pi\right]\:\:\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\Rightarrow\:\:\:\:\theta=\:\frac{\pi}{\mathrm{6}}\:;\:\mathrm{and}\:\theta=\frac{\mathrm{5}\pi}{\mathrm{6}}\:. \\ $$
Commented by mrW1 last updated on 07/Jul/17
$$\mathrm{I}\:\mathrm{think}\:\theta=\frac{\pi}{\mathrm{6}}\:\mathrm{and}\:\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$
Commented by Tinkutara last updated on 08/Jul/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$