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Question Number 135204 by victoras last updated on 11/Mar/21

Answered by benjo_mathlover last updated on 11/Mar/21

81^(sin^2 x) + 81^(1−sin^2 x) = 30 let 81^(sin^2 x) = z ⇒z + ((81)/z) = 30 ⇒z^2 −30z +81 = 0 ⇒(z−27)(z−3)=0 { ((81^(sin^2 x) = 27 ⇒sin^2 x = (3/4))),((81^(sin^2 x) = 3 ⇒sin^2 x = (1/4))) :}

$$\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:+\:\mathrm{81}^{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:=\:\mathrm{30} \\ $$$$\mathrm{let}\:\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:=\:\mathrm{z}\: \\ $$$$\Rightarrow\mathrm{z}\:+\:\frac{\mathrm{81}}{\mathrm{z}}\:=\:\mathrm{30}\: \\ $$$$\Rightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{30z}\:+\mathrm{81}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{z}−\mathrm{27}\right)\left(\mathrm{z}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:=\:\mathrm{27}\:\Rightarrow\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\frac{\mathrm{3}}{\mathrm{4}}}\\{\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:=\:\mathrm{3}\:\Rightarrow\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$

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