InterviewSolution
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InterviewSolution
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Solve: `16^sin^(2x)16^cos^(2x)=10 ,0lt=x |
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Answer» `16^(sin^(2)x)+16^(1-sin^(2) x)=10` If `16^(sin^(2)x)=t`, then `t+16/t=10` Then Eq. (i) becomes `t^(2)-10t+16=0` or `t=2, 8` `rArr 16^(sin^(2) x)=16^(1//4)` or `16^(3//4)` `rArr sin x= pm 1/2, pm sqrt(3)/2` Now `sin x=1/2`, then `x=pi/6, (5pi)/6` `sin x=-1/2`, then `x=(7pi)/6` or `(11pi)/6` `sin x= sqrt(3)/2`, then `x=pi/3, (2pi)/3` `sin x= - sqrt(3)/2`, then `x=(4pi)/3, (5 pi)/3` Hence, there will be eight solutions in all. |
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