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Question Number 33116 by abdo imad last updated on 10/Apr/18

solve at [0,π]  cosα +cos(2α) +cos(3α)=0

$${solve}\:{at}\:\left[\mathrm{0},\pi\right]\:\:{cos}\alpha\:+{cos}\left(\mathrm{2}\alpha\right)\:+{cos}\left(\mathrm{3}\alpha\right)=\mathrm{0} \\ $$

Answered by MJS last updated on 10/Apr/18

cos 2α=2cos^2 α−1  cos 3α=4cos^3 α−3cos α  cos α+2cos^2 α−1+4cos^3 α−3cos α=0  4cos^3 α+2cos^2 α−2cos α−1=0  cos^3 α+(1/2)cos^2 α−(1/2)cos α−(1/4)=0  trying cos α=±(1/4); ±(1/2) ⇒  ⇒ (cos α)_1 =−(1/2) ⇒ α_1 =((2π)/3)  ((cos^3 α+(1/2)cos^2 α−(1/2)cos α−(1/4))/(cos α+(1/2)))=0  cos^2 α−(1/2)=0  (cos α)_2 =−((√2)/2) ⇒ α_2 =((3π)/4)  (cos α)_3 =((√2)/2) ⇒ α_3 =(π/4)

$$\mathrm{cos}\:\mathrm{2}\alpha=\mathrm{2cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{3}\alpha=\mathrm{4cos}^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha \\ $$$$\mathrm{cos}\:\alpha+\mathrm{2cos}^{\mathrm{2}} \alpha−\mathrm{1}+\mathrm{4cos}^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha=\mathrm{0} \\ $$$$\mathrm{4cos}^{\mathrm{3}} \alpha+\mathrm{2cos}^{\mathrm{2}} \alpha−\mathrm{2cos}\:\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{3}} \alpha+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \alpha−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\alpha−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{cos}\:\alpha=\pm\frac{\mathrm{1}}{\mathrm{4}};\:\pm\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:\left(\mathrm{cos}\:\alpha\right)_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\alpha_{\mathrm{1}} =\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\frac{\mathrm{cos}^{\mathrm{3}} \alpha+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}^{\mathrm{2}} \alpha−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\alpha−\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{cos}\:\alpha+\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{0} \\ $$$$\mathrm{cos}^{\mathrm{2}} \alpha−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left(\mathrm{cos}\:\alpha\right)_{\mathrm{2}} =−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\:\alpha_{\mathrm{2}} =\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$$\left(\mathrm{cos}\:\alpha\right)_{\mathrm{3}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\:\alpha_{\mathrm{3}} =\frac{\pi}{\mathrm{4}} \\ $$

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