Question Number 17151 by Tinkutara last updated on 01/Jul/17
![The solution of the equation cos^2 θ − 2cosθ = 4sinθ − sin2θ where θ ∈ [0, π] is](https://www.tinkutara.com/question/Q17151.png)
$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta\:−\:\mathrm{2cos}\theta\:=\:\mathrm{4sin}\theta\:−\:\mathrm{sin2}\theta\:\mathrm{where} \\ $$$$\theta\:\in\:\left[\mathrm{0},\:\pi\right]\:\mathrm{is} \\ $$
Answered by ajfour last updated on 01/Jul/17
![4sin θ−2sin θcos θ+2cos θ−cos^2 θ=0 2sin θ(2−cos θ)+cos θ(2−cos θ)=0 (2−cos θ)(2sin θ+cos θ)=0 ⇒ 2sin θ+cos θ=0 or tan θ=−(1/2) In [0,π], θ=π−tan^(−1) ((1/2)) .](https://www.tinkutara.com/question/Q17154.png)
$$\mathrm{4sin}\:\theta−\mathrm{2sin}\:\theta\mathrm{cos}\:\theta+\mathrm{2cos}\:\theta−\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\mathrm{2sin}\:\theta\left(\mathrm{2}−\mathrm{cos}\:\theta\right)+\mathrm{cos}\:\theta\left(\mathrm{2}−\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}−\mathrm{cos}\:\theta\right)\left(\mathrm{2sin}\:\theta+\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{2sin}\:\theta+\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\mathrm{or}\:\:\:\:\mathrm{tan}\:\theta=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{In}\:\left[\mathrm{0},\pi\right],\:\:\theta=\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:. \\ $$
Commented by Tinkutara last updated on 01/Jul/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$