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Question Number 10836 by okhema last updated on 27/Feb/17
solve cos2θ−3cosθ=1  for o≤θ≤2π
$${solve}\:{cos}\mathrm{2}\theta−\mathrm{3}{cos}\theta=\mathrm{1} \\ $$$${for}\:{o}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$
Answered by ridwan balatif last updated on 27/Feb/17
cos2θ−3cosθ=1  (2cos^2 θ−1)−3cosθ=1  2cos^2 θ−3cosθ−2=0  (2cosθ+1)(cosθ−2)=0  cosθ=−1/2 ∨ cosθ=2(IMPOSSIBLE)    cosθ=−1/2  θ=(2/3)π,(4/3)π
$$\mathrm{cos2}\theta−\mathrm{3cos}\theta=\mathrm{1} \\ $$$$\left(\mathrm{2cos}^{\mathrm{2}} \theta−\mathrm{1}\right)−\mathrm{3cos}\theta=\mathrm{1} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \theta−\mathrm{3cos}\theta−\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{2cos}\theta+\mathrm{1}\right)\left(\mathrm{cos}\theta−\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{cos}\theta=−\mathrm{1}/\mathrm{2}\:\vee\:\mathrm{cos}\theta=\mathrm{2}\left(\mathrm{IMPOSSIBLE}\right) \\ $$$$ \\ $$$$\mathrm{cos}\theta=−\mathrm{1}/\mathrm{2} \\ $$$$\theta=\frac{\mathrm{2}}{\mathrm{3}}\pi,\frac{\mathrm{4}}{\mathrm{3}}\pi \\ $$

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