The question can be rewriten as:
[tex]2\sin ^2(\theta)=1[/tex]First, we isolate the sin(θ):
[tex]\sin ^2\theta=\frac{1}{2}[/tex]On the left side we have a value of [sin(Θ) ]². The first thing to do, so we isolate Θ, we must calculate the square root in both sides. It results in the following:
[tex]\begin{gathered} \sqrt[]{\sin^2\theta}=\sqrt{\frac{1}{2}} \\ \sin \theta=\pm\sqrt[]{\frac{1}{2}}=\pm\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]If we apply the inverse function (sin⁻¹) in both sides, we can do as follow:
[tex]\begin{gathered} \sin ^{-1}(\sin (\theta))=\sin ^{-1}(\pm\frac{\sqrt[]{2}}{2}) \\ \theta=\sin ^{-1}(\pm\frac{\sqrt[]{2}}{2}) \end{gathered}[/tex]If we use ta unitary circle to better look at this, we can find two values of θ for each sine of θ. It is:
For the positive value of sin(θ), we find:
[tex]\begin{gathered} \theta=45\degree=\frac{\pi}{4} \\ \theta=135\degree=\frac{3\pi}{4} \end{gathered}[/tex]For the negative value of sin(θ), we find:
[tex]\begin{gathered} \theta=225\degree=\frac{5\pi}{4} \\ \theta=315\degree=\frac{7\pi}{4} \end{gathered}[/tex]Because θ is from 0 to 2π, all this values satisfy the trigonometric equation. If you task is to find a single value for θ, you may use the fist one: π/4. Because it is more usual to work with sine with θ in the range of -π/2 to π/2, and the positive value is preferable.
