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Question Number 129070 by benjo_mathlover last updated on 12/Jan/21

Solve 2cos x cos 2x = cos ((π/3) −3x)

$$\:\mathrm{Solve}\:\mathrm{2cos}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}\:−\mathrm{3x}\right) \\ $$

Answered by liberty last updated on 12/Jan/21

⇔ 2cos x cos 2x = cos ((π/3)−3x) ⇔ cos 3x+cos x = (1/2)cos 3x+((√3)/2) sin 3x ⇔ cos x = ((√3)/2) sin 3x−(1/2)cos 3x ⇔ cos x = sin (3x−(π/6)) ⇔ sin ((π/2)−x)=sin (3x−(π/6)) ⇔ x = { ((((3k+1)π)/6)),((((3k+1)π)/3)) :}

$$\:\Leftrightarrow\:\mathrm{2cos}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:=\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{3x}\right) \\ $$$$\:\Leftrightarrow\:\mathrm{cos}\:\mathrm{3x}+\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{3x}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{3x} \\ $$$$\:\Leftrightarrow\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{3x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{3x}\: \\ $$$$\:\Leftrightarrow\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{sin}\:\left(\mathrm{3x}−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\:\Leftrightarrow\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)=\mathrm{sin}\:\left(\mathrm{3x}−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\:\Leftrightarrow\:\mathrm{x}\:=\:\begin{cases}{\frac{\left(\mathrm{3k}+\mathrm{1}\right)\pi}{\mathrm{6}}}\\{\frac{\left(\mathrm{3k}+\mathrm{1}\right)\pi}{\mathrm{3}}}\end{cases} \\ $$

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