[tex]\cos (2t)-2\sin ^2(t)=0[/tex]
Use the next trigonometric rules:
[tex]\begin{gathered} \cos 2t=\cos ^2t-\sin ^2t \\ \\ \sin ^2t=1-\cos ^2t \end{gathered}[/tex]
Use Cos2t
[tex]\cos ^2t-\sin ^2t-2\sin ^2t=0[/tex]
Combine similar terms:
[tex]\cos ^2t-3\sin ^2t=0[/tex]
Use sin²t:
[tex]\begin{gathered} \cos ^2t-3(1-\cos ^2t)=0 \\ \\ \cos ^2t-3+3\cos ^2t=0 \end{gathered}[/tex]
Combine similar terms:
[tex]4\cos ^2t-3=0[/tex]
Add 3 in both sides of the equation:
[tex]\begin{gathered} 4\cos ^2t-3+3=0+3 \\ 4\cos ^2t=3 \end{gathered}[/tex]
Divide both sides of the equation into 4:
[tex]\begin{gathered} \frac{4\cos ^2t}{4}=\frac{3}{4} \\ \\ \cos ^2t=\frac{3}{4} \end{gathered}[/tex]
Find the square root of both sides of the equation:
[tex]\begin{gathered} \sqrt[]{\cos^2t}=\sqrt[]{\frac{3}{4}} \\ \\ \cos t=\pm\frac{\sqrt[]{3}}{2} \\ \\ \cos t=+\frac{\sqrt[]{3}}{2} \\ \\ \cos t=-\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]
Use the unit circle to find wich angles in the given interval have a cos equal to:
[tex]\cos t=\pm\frac{\sqrt[]{3}}{2}[/tex]
Solution:
[tex]t=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}[/tex]
