Question

The solution of trigonometric equation ${\cos }^4\left(x\right)+{\sin }^4\left(x\right)=2\cos (2x+\mathrm{\pi })\cos (2\mathrm{x}-\mathrm{\pi })$ is

• A. $x=\frac{n\mathrm{\pi }}{2}\pm {\sin }^{-1}\left(\frac{1}{5}\right)$
• B. $x=\frac{n\mathrm{\pi }}{4}+\frac{{\left(-1\right)}^n}{4}{\sin }^{-1}\left(\frac{\pm 2\sqrt{2}}{3}\right)$
• C. $x=\frac{n\mathrm{\pi }}{2}\pm {\cos }^{-1}\left(\frac{1}{5}\right)$
• D. $x=\frac{n\mathrm{\pi }}{2}+\frac{{\left(-1\right)}^n}{4}{\cos }^{-1}\left(\frac{1}{5}\right)$

Answer 1

The correct option is B

$x=\frac{n\mathrm{\pi }}{4}+\frac{{\left(-1\right)}^n}{4}{\sin }^{-1}\left(\frac{\pm 2\sqrt{2}}{3}\right)$

Explanation for correct option:

Given: ${\cos }^4\left(x\right)+{\sin }^4\left(x\right)=2\cos (2x+\mathrm{\pi })\cos (2\mathrm{x}-\mathrm{\pi })$

$$\begin{gathered} \Rightarrow {\left({\cos }^2\left(x\right)\right)}^2+{\left({\sin }^2\left(x\right)\right)}^2=\cos (2x+\mathrm{\pi }+2\mathrm{x}+\mathrm{\pi })+\cos (2x+\mathrm{\pi }-2\mathrm{x}+\mathrm{\pi })\mathbf{\left[}\mathbf{\because }\mathbf{2}\mathbf{cos}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{=}\mathbf{cos}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{+}\mathbf{cos}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{\right]} \\ \Rightarrow {\left({\cos }^2\left(x\right)+{\sin }^2\left(x\right)\right)}^2-\left(2{\sin }^2\right(x\left){\cos }^2\right(x\left)\right)=\cos (4x+2\mathrm{\pi })+\cos \left(2\mathrm{\pi }\right)\mathbf{[}\mathbf{\because }{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{=}{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{ab}\mathbf{]} \\ \Rightarrow 1-\frac{{\left(2\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\right)}^2}{2}=\cos \left(4\mathrm{x}\right)+1\mathbf{\left[}\mathbf{\because }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{2}\mathbf{\pi }\mathbf{+}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{,}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{2}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{\right]} \\ \Rightarrow -\frac{{\sin }^2\left(2\mathrm{x}\right)}{2}=\cos \left(4\mathrm{x}\right) \\ \Rightarrow -\frac{1-\cos \left(4\mathrm{x}\right)}{2\times 2}=\cos \left(4\mathrm{x}\right)\mathbf{\left[}\mathbf{\because }\mathbf{cos}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]} \\ \Rightarrow -1+\cos \left(4\mathrm{x}\right)=4\cos \left(4\mathrm{x}\right) \\ \Rightarrow 3\cos \left(4\mathrm{x}\right)=-1 \\ \Rightarrow \cos \left(4\mathrm{x}\right)=-\frac{1}{3} \\ \Rightarrow \sin \left(4\mathrm{x}\right)=\sqrt{1-\left({\cos }^2\right(4\mathrm{x}\left)\right)} \\ \Rightarrow \sin \left(4\mathrm{x}\right)=\sqrt{1-{\left(\frac{1}{3}\right)}^2} \\ \Rightarrow \sin \left(4\mathrm{x}\right)=\sqrt{1-\frac{1}{9}} \\ \Rightarrow \sin \left(4\mathrm{x}\right)=\pm \frac{2\sqrt{2}}{3} \end{gathered}$$

so, $x=\left(\frac{\mathrm{n\pi }}{4}\right)+\frac{{\left(-1\right)}^n}{4}{\sin }^{-1}\left(\frac{\pm 2\sqrt{2}}{3}\right)\mathbf{\left[}\mathbf{\because }\mathbf{}\mathbf{sin}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{\left(}\mathbf{n\pi }\mathbf{\right)}\mathbf{+}{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{\right]}$

Hence, option B is correct.