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True and False
One value of θ which satisfies the equation \sin^4 \theta - 2\sin^2 \theta - 1 lies between 0 and 2π

 

Answers (1)

Given equation is  \sin ^{4} \theta - 2\sin ^{2} \theta - 1=0 \\\\


  \\ \sin ^{2} \theta =\frac{ - \left( - 2 \right) \pm \sqrt { \left( - 2 \right) ^{2} - 4 \times 1 \times \left( - 1 \right) }}{2 \times 1}=\frac{2 \pm \sqrt {4+4}}{2} \\\\ =\frac{2 \pm \sqrt {8}}{2}=\frac{2 \pm 2\sqrt {2}}{2}=1 \pm \sqrt {2} \\\\

     - 1 \leq \sin \theta \leq 1~ and \sin ^{2} \theta \leq 1~ but \sin ^{2} \theta =1 \pm \sqrt {2}

which is not possible

Hence, the given statement is ‘false’   

Posted by

infoexpert21

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