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Question Number 164901 by cortano1 last updated on 23/Jan/22

	Answered by bobhans last updated on 23/Jan/22
	$⇒ ((sin^4 x)/(cos^4 x)) −((2−3cos^2 x)/(cos^2 x)) = 0 ⇒ (((1−c)^2 )/c^2 ) −((2−3c)/c) = 0 , c = cos^2 x ⇒(1−c)^2 −c(2−3c)=0 ⇒c^2 −2c+1−2c+3c^2 = 0 ⇒4c^2 −4c+1= 0 ⇒(2c−1)^2 =0 ⇒ { ((cos x=(1/2)(√2))),((cos x=−(1/2)(√2))) :}⇒ x=(π/4); ((3π)/4); ((5π)/4); ((7π)/4)$
	$\Rightarrow \frac{\sin^{4} x}{\cos^{4} x} - \frac{2 - 3 \cos^{2} x}{\cos^{2} x} = 0$ $\Rightarrow \frac{{(1 - c)}^{2}}{c^{2}} - \frac{2 - 3 c}{c} = 0, c = \cos^{2} x$ $\Rightarrow {(1 - c)}^{2} - c (2 - 3 c) = 0$ $\Rightarrow c^{2} - 2 c + 1 - 2 c + {3 c}^{2} = 0$ $\Rightarrow {4 c}^{2} - 4 c + 1 = 0$ $\Rightarrow {(2 c - 1)}^{2} = 0$ $\Rightarrow {\begin{cases} \cos x = \frac{1}{2} \sqrt{2} \\ \cos x = - \frac{1}{2} \sqrt{2} \end{cases} \Rightarrow x = \frac{π}{4}; \frac{3 π}{4}; \frac{5 π}{4}; \frac{7 π}{4}$
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