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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)

$$\:\Rightarrow\:\frac{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\:−\frac{\mathrm{2}−\mathrm{3cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{1}−\mathrm{c}\right)^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }\:−\frac{\mathrm{2}−\mathrm{3c}}{\mathrm{c}}\:=\:\mathrm{0}\:,\:\mathrm{c}\:=\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{c}\right)^{\mathrm{2}} −\mathrm{c}\left(\mathrm{2}−\mathrm{3c}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{c}^{\mathrm{2}} −\mathrm{2c}+\mathrm{1}−\mathrm{2c}+\mathrm{3c}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{4c}^{\mathrm{2}} −\mathrm{4c}+\mathrm{1}=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2c}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{cos}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}}\\{\mathrm{cos}\:\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}}\end{cases}\Rightarrow\:\mathrm{x}=\frac{\pi}{\mathrm{4}};\:\frac{\mathrm{3}\pi}{\mathrm{4}};\:\frac{\mathrm{5}\pi}{\mathrm{4}};\:\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$

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