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Question Number 131661 by benjo_mathlover last updated on 07/Feb/21

log _((√2) sin x) (1+cos x) = 2 −((2π)/3)≤x≤(π/3)

$$\:\:\:\mathrm{log}\:_{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)\:=\:\mathrm{2} \\ $$$$\:−\frac{\mathrm{2}\pi}{\mathrm{3}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{3}} \\ $$

Answered by liberty last updated on 07/Feb/21

{ (((√2) sin x >0 ; x in I or II quadrant)),(((√2) sin x ≠ 1⇒sin x≠(1/( (√2))))) :} ⇔ 1+cos x = ((√2) sin x)^2 ⇔2sin^2 x−cos x−1=0 ⇔2−2cos^2 x−cos x−1=0 ⇔2cos^2 x+cos x−1=0 ⇔(2cos x−1)(cos x+1)=0 { ((cos x=(1/2)⇒x = (π/3)←only solution)),((cos x=−1 (reject))) :}

$$\:\begin{cases}{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\:>\mathrm{0}\:;\:\mathrm{x}\:\mathrm{in}\:\mathrm{I}\:\mathrm{or}\:\mathrm{II}\:\mathrm{quadrant}}\\{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\:\neq\:\mathrm{1}\Rightarrow\mathrm{sin}\:\mathrm{x}\neq\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\end{cases} \\ $$$$\Leftrightarrow\:\mathrm{1}+\mathrm{cos}\:\mathrm{x}\:=\:\left(\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{2cos}\:\mathrm{x}−\mathrm{1}\right)\left(\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}\:=\:\frac{\pi}{\mathrm{3}}\leftarrow\mathrm{only}\:\mathrm{solution}}\\{\mathrm{cos}\:\mathrm{x}=−\mathrm{1}\:\left(\mathrm{reject}\right)}\end{cases} \\ $$

Commented by liberty last updated on 07/Feb/21

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