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Question Number 136121 by bramlexs22 last updated on 18/Mar/21
3 cos x = 13 sin (((2x)/3)) + 17 cos ((x/3))
$$\mathrm{3}\:\mathrm{cos}\:{x}\:=\:\mathrm{13}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\:+\:\mathrm{17}\:\mathrm{cos}\:\left(\frac{{x}}{\mathrm{3}}\right) \\ $$
Answered by EDWIN88 last updated on 19/Mar/21
let (x/3) = t ⇒3cos 3t = 13sin 2t+17cos t ⇒3(4cos^3 t−3cos t)=26sin t cos t +17cos t 12cos^3 t−26cos t−26sin t cos t = 0 2cos t (6cos^2 t−13sin t−13)=0 (∗) 2cos t = 0 ⇒t=±(π/2)+2nπ (x/3) = ± (π/2)+2nπ ⇒x = ± ((3π)/2)+6nπ (∗)⇒6−6sin^2 t−13sin t−13 =0 ⇒ 6sin^2 t+13sin t+7=0 ⇒(6sin t+7)(sin t+1)=0 { ((sin t =−(7/6) ⇒t∉R)),((sin t=−1⇒t=−(π/2)+2nπ , (x/3)=−(π/2)+2nπ)) :}
$$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{3}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{3cos}\:\mathrm{3t}\:=\:\mathrm{13sin}\:\mathrm{2t}+\mathrm{17cos}\:\mathrm{t} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{4cos}\:^{\mathrm{3}} \mathrm{t}−\mathrm{3cos}\:\mathrm{t}\right)=\mathrm{26sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:+\mathrm{17cos}\:\mathrm{t} \\ $$$$\mathrm{12cos}\:^{\mathrm{3}} \mathrm{t}−\mathrm{26cos}\:\mathrm{t}−\mathrm{26sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:=\:\mathrm{0} \\ $$$$\mathrm{2cos}\:\mathrm{t}\:\left(\mathrm{6cos}\:^{\mathrm{2}} \mathrm{t}−\mathrm{13sin}\:\mathrm{t}−\mathrm{13}\right)=\mathrm{0} \\ $$$$\left(\ast\right)\:\mathrm{2cos}\:\mathrm{t}\:=\:\mathrm{0}\:\Rightarrow\mathrm{t}=\pm\frac{\pi}{\mathrm{2}}+\mathrm{2n}\pi \\ $$$$\:\frac{\mathrm{x}}{\mathrm{3}}\:=\:\pm\:\frac{\pi}{\mathrm{2}}+\mathrm{2n}\pi\:\Rightarrow\mathrm{x}\:=\:\pm\:\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{6n}\pi \\ $$$$\left(\ast\right)\Rightarrow\mathrm{6}−\mathrm{6sin}\:^{\mathrm{2}} \mathrm{t}−\mathrm{13sin}\:\mathrm{t}−\mathrm{13}\:=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{6sin}\:^{\mathrm{2}} \mathrm{t}+\mathrm{13sin}\:\mathrm{t}+\mathrm{7}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{6sin}\:\mathrm{t}+\mathrm{7}\right)\left(\mathrm{sin}\:\mathrm{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{sin}\:\mathrm{t}\:=−\frac{\mathrm{7}}{\mathrm{6}}\:\Rightarrow\mathrm{t}\notin\mathbb{R}}\\{\mathrm{sin}\:\mathrm{t}=−\mathrm{1}\Rightarrow\mathrm{t}=−\frac{\pi}{\mathrm{2}}+\mathrm{2n}\pi\:,\:\frac{\mathrm{x}}{\mathrm{3}}=−\frac{\pi}{\mathrm{2}}+\mathrm{2n}\pi}\end{cases} \\ $$
Commented by EDWIN88 last updated on 19/Mar/21
yes. thanks you
$$\mathrm{yes}.\:\mathrm{thanks}\:\mathrm{you} \\ $$

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