Question

solve, tan^2. x. --2sinx = 0​

Answer 1

$\huge \mathbb {Answer:-}$

_________

We have,

$\implies tan^2\: x - 2 \: sin \: x = 0$

$\implies ( sin^2 \: x / cos^2 \: x) - 2 \: sin \: x = 0$

$\implies ( sin^2 \: x - 2 \: sin \: x \: cos^2 \: x) / cos^2 \: x = 0$

$\implies sin \: x ( sin \: x - 2\:cos^2 x) = 0$

$\implies sin \: x = 2\: cos^2 \: x$

$\implies sin \: x = 2 - 2 \: sin^2 \: x$

$\implies 2 \: sin^2 \:x + sin \: x - 2 = 0$

Therefore, we get :

$sin \: x = [- ( 1) + </p><p>\sqrt{ 1^2 - 4. 2. (-2)}] / 4$

$= ( - 1 + \sqrt{17} ) / 4$

Thus :

$x = sin^{-1} ( - 1 + \sqrt {17} ) / 4$

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