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Question Number 113824 by deepraj123 last updated on 15/Sep/20

If 4 sin^(−1) x+cos^(−1) x= π, then x equals

$$\mathrm{If}\:\mathrm{4}\:\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{cos}^{−\mathrm{1}} {x}=\:\pi,\:\mathrm{then}\:{x}\:\mathrm{equals} \\ $$

Answered by $@y@m last updated on 15/Sep/20

Let sin^(−1) x=α ⇒sin α=x ...(1)  Let cos ^(−1) x=β ⇒ cos β=x ...(2)  ATQ,  4α+β=π  4α=π−β  cos  4α=cos  (π−β)  cos  4α=−cos β  cos  4α=−sin α  1−sin^2 2α=−sin α  sin^2 2α=1+sin α  (2sin αcos α)^2 =1+sin α  (2x(√(1−x^2 )))^2 =1+x  4x^2 (1−x^2 )=1+x  4x^2 (1+x)(1−x)=1+x  (1+x)(4x^2 −4x^3 −1)=0  x=−1

$${Let}\:\mathrm{sin}\:^{−\mathrm{1}} {x}=\alpha\:\Rightarrow\mathrm{sin}\:\alpha={x}\:...\left(\mathrm{1}\right) \\ $$$${Let}\:\mathrm{cos}\:\:^{−\mathrm{1}} {x}=\beta\:\Rightarrow\:\mathrm{cos}\:\beta={x}\:...\left(\mathrm{2}\right) \\ $$$${ATQ}, \\ $$$$\mathrm{4}\alpha+\beta=\pi \\ $$$$\mathrm{4}\alpha=\pi−\beta \\ $$$$\mathrm{cos}\:\:\mathrm{4}\alpha=\mathrm{cos}\:\:\left(\pi−\beta\right) \\ $$$$\mathrm{cos}\:\:\mathrm{4}\alpha=−\mathrm{cos}\:\beta \\ $$$$\mathrm{cos}\:\:\mathrm{4}\alpha=−\mathrm{sin}\:\alpha \\ $$$$\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\alpha=−\mathrm{sin}\:\alpha \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\alpha=\mathrm{1}+\mathrm{sin}\:\alpha \\ $$$$\left(\mathrm{2sin}\:\alpha\mathrm{cos}\:\alpha\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{sin}\:\alpha \\ $$$$\left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1}+{x} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)=\mathrm{1}+{x} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)=\mathrm{1}+{x} \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{1} \\ $$

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