Answer:
[tex]x=\frac{\pi}{3},\frac{5\pi}{3},\pi[/tex]Explanation:
Given the below;
[tex]2\sin ^2x=1+\cos x[/tex]We'll follow the below steps to solve for x;
Step 1: Rewrite using the below trig identity;
[tex]\begin{gathered} \sin ^2x+\cos ^2x=1 \\ \therefore\sin ^2x=1-\cos ^2x \end{gathered}[/tex]So, we'll have;
[tex]\begin{gathered} 2(1-\cos ^2x)=1+\cos x \\ \end{gathered}[/tex]Step 2: Apply the difference of squares formula to the left-hand side of the equation;
[tex]\begin{gathered} 2(1^2-\cos ^2x)=1+\cos x \\ 2(1+\cos x)(1-\cos x)=1+\cos x \end{gathered}[/tex]Step 3: Subtract (1 + cos x) from both sides;
[tex]2(1+\cos x)(1-\cos x)-(1+\cos x)=0[/tex]Step 4: Factor out (1 + cos x);
[tex]\begin{gathered} (1+\cos x)\lbrack2(1-\cos x)-1\rbrack=0 \\ (1+\cos x)(2-2\cos x-1)=0_{} \\ (1+\cos x)(1-2\cos x)=0 \end{gathered}[/tex]Step 5: Solve for the values of x by equating each term to zero;
[tex]\begin{gathered} 1+\cos x=0 \\ \cos x=-1 \\ x=\cos ^{-1}(-1) \\ x=180^{\circ} \\ \therefore x=\pi \\ Or \\ 1-2\cos x=0 \\ -2\cos x=-1 \\ \frac{-2\cos x}{-2}=\frac{-1}{-2} \\ \cos x=\frac{1}{2} \\ x=\cos ^{-1}(\frac{1}{2}) \\ x=60^{\circ},300^{\circ} \\ \therefore x=\frac{\pi}{3},\frac{5\pi}{3} \end{gathered}[/tex][tex]\begin{gathered} \\ \end{gathered}[/tex]
