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Question Number 80739 by jagoll last updated on 05/Feb/20

cos^3 θ+2sin^2 θ=3 ^� θ∈(0,2π)  what is θ ?

$$\mathrm{cos}\:^{\mathrm{3}} \theta+\mathrm{2sin}\:^{\mathrm{2}} \theta=\mathrm{3}\bar {\:}\theta\in\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$$${what}\:{is}\:\theta\:? \\ $$

Commented by mr W last updated on 06/Feb/20

no solution! because LHS max=2<RJS!

$${no}\:{solution}!\:{because}\:{LHS}\:{max}=\mathrm{2}<{RJS}! \\ $$

Commented by hknkrc46 last updated on 06/Feb/20

sin^2 θ=1−cos^2 θ  cos^3 θ+2(1−cos^2 θ)=3  cos^3 θ+2−2cos^2 θ=3  cos^3 θ−2cos^2 θ=1  cos^2 θ(cosθ−2)=1  −1≤cosθ≤1⇒−3≤cosθ−2≤−1  ⇒cos^2 θ<0⇒θ∉(0,2π)

$${sin}^{\mathrm{2}} \theta=\mathrm{1}−{cos}^{\mathrm{2}} \theta \\ $$$${cos}^{\mathrm{3}} \theta+\mathrm{2}\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)=\mathrm{3} \\ $$$${cos}^{\mathrm{3}} \theta+\mathrm{2}−\mathrm{2}{cos}^{\mathrm{2}} \theta=\mathrm{3} \\ $$$${cos}^{\mathrm{3}} \theta−\mathrm{2}{cos}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$${cos}^{\mathrm{2}} \theta\left({cos}\theta−\mathrm{2}\right)=\mathrm{1} \\ $$$$−\mathrm{1}\leqslant{cos}\theta\leqslant\mathrm{1}\Rightarrow−\mathrm{3}\leqslant{cos}\theta−\mathrm{2}\leqslant−\mathrm{1} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} \theta<\mathrm{0}\Rightarrow\theta\notin\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$

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