SOLUTION
Step 1 :
In this question, we are meant to find all the solutions of :
[tex]\cos \text{ 2 x = sin x in the interval }\lbrack\text{ 0, 2}\pi\text{ )}[/tex][tex]\begin{gathered} \text{Given that } \\ \cos 2x=cos^2x-sin^2x \end{gathered}[/tex][tex]\begin{gathered} \cos ^2x-sin^2x\text{ = sin x} \\ (1-sin^2x)-sin^2x\text{ = sin x} \\ \text{Let p = sinx} \\ (1-p^2)-p^2\text{ = p} \\ 1-2p^2\text{ = p} \\ 2p^2\text{ + p - 1= 0} \end{gathered}[/tex]Step 2 :
Factorising
[tex]\begin{gathered} 2p^2\text{ + p - 1 = 0} \\ 2p^2\text{ + 2p - p - 1 = 0} \\ 2p\text{ ( p + 1 ) - 1 ( p + 1 ) = 0} \\ (\text{ p + 1 ) ( 2p - 1 ) = 0} \\ \text{set p + 1 = 0 or 2p - 1 = 0} \\ p\text{ = 0 - 1 or 2p = 1} \\ p\text{ = - 1 or p =}\frac{1}{2} \end{gathered}[/tex]Step 3 :
[tex]\begin{gathered} \sin ce\text{ p = sin x} \\ \text{Case 1 : } \\ p\text{ = sin x = - 1} \\ x\text{ = }\sin ^{-1}\text{ ( -1 )} \\ \text{x = -90}^{0\text{ }}(+360^0\text{ ) ( In the interval of }\lbrack\text{ 0 , 2}\pi\text{ )} \\ x=270^0 \\ \text{x = }\frac{3\text{ }\pi}{4} \end{gathered}[/tex]Step 4 :
[tex]\begin{gathered} \text{Case 2 :} \\ p\text{ = }\frac{1}{2\text{ }}\text{ = sinx } \\ x\text{ = }\sin ^{-1}\text{ ( }\frac{1}{2\text{ }}\text{ )} \\ x=30^0^{} \\ \text{x = }\frac{\pi}{12} \end{gathered}[/tex]Step 5 :
[tex]\begin{gathered} \text{Case 3 :} \\ \sin \text{ x = p =}\frac{1}{2} \\ \sin x\text{ = }\frac{1}{2} \\ x\text{ = }\sin ^{-1}(\text{ }\frac{1}{2})=150^0 \\ x\text{ =}\frac{5\pi}{12} \end{gathered}[/tex]CONCLUSION :
The values of x are:
[tex]\begin{gathered} x\text{ = }\frac{3\pi}{4} \\ x\text{ = }\frac{\pi}{12} \\ x\text{ = }\frac{5\pi}{12} \end{gathered}[/tex]
