Question

If $2{\sin }^2\beta -{\cos }^2\beta =2$the $\beta$ is.

Answer 1

Find the required value:

Given : $2{\sin }^2\beta -{\cos }^2\beta =2$

we know that ${\sin }^2\beta +{\cos }^2\beta =1$

Thus, ${\cos }^2\beta =1-{\sin }^2\beta$

Substitute the value in the given equation.

$$\begin{gathered} 2{\sin }^2\beta -1+{\sin }^2\beta =2 \\ \Rightarrow 3{\sin }^2\beta =3 \\ \Rightarrow {\sin }^2\beta =1 \\ \Rightarrow \sin \beta =1 \\ \Rightarrow \beta =90°(0\le \beta \le 360) \end{gathered}$$

Hence, the value of $\beta$ is $90°,(0\le \beta \le 360)$