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Question Number 116800 by mr W last updated on 06/Oct/20

Answered by bobhans last updated on 07/Oct/20

⇒((sin 8x)/(sin 4x)) = ((sin 5x)/(sin 3x)) ; ((2sin 4x.cos 4x)/(sin 4x)) = ((sin 5x)/(sin 3x)) ⇒2sin 3x.cos 4x = sin 5x ⇒sin 7x+sin (−x)=sin 5x ⇒sin 7x−sin 5x +sin (−x)=0 ⇒2cos 6x.sin x−sin x = 0 { ((2cos 6x = 1 ⇒cos 6x = (1/2); x = 10°)),((sin x = 0 ← rejected)) :}

$$\Rightarrow\frac{\mathrm{sin}\:\mathrm{8x}}{\mathrm{sin}\:\mathrm{4x}}\:=\:\frac{\mathrm{sin}\:\mathrm{5x}}{\mathrm{sin}\:\mathrm{3x}}\:;\:\frac{\mathrm{2sin}\:\mathrm{4x}.\mathrm{cos}\:\mathrm{4x}}{\mathrm{sin}\:\mathrm{4x}}\:=\:\frac{\mathrm{sin}\:\mathrm{5x}}{\mathrm{sin}\:\mathrm{3x}} \\ $$$$\Rightarrow\mathrm{2sin}\:\mathrm{3x}.\mathrm{cos}\:\mathrm{4x}\:=\:\mathrm{sin}\:\mathrm{5x} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{7x}+\mathrm{sin}\:\left(−\mathrm{x}\right)=\mathrm{sin}\:\mathrm{5x} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{7x}−\mathrm{sin}\:\mathrm{5x}\:+\mathrm{sin}\:\left(−\mathrm{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2cos}\:\mathrm{6x}.\mathrm{sin}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{2cos}\:\mathrm{6x}\:=\:\mathrm{1}\:\Rightarrow\mathrm{cos}\:\mathrm{6x}\:=\:\frac{\mathrm{1}}{\mathrm{2}};\:\mathrm{x}\:=\:\mathrm{10}°}\\{\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{0}\:\leftarrow\:\mathrm{rejected}}\end{cases} \\ $$

Commented by mr W last updated on 07/Oct/20

nice!

$${nice}! \\ $$

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