Question

solve sin² theta + sin theta = 1 and find theta​

Answer 1

The value of $\theta=\sin^{-1}(\frac{-1\pm \sqrt5}{2})$

Step-by-step explanation:

Given:

The equation to solve is given as:

$\sin^2 \theta+\sin \theta=1$

Let $\sin \theta=x$. So, the equation becomes:

$x^2 +x=1\\x^2+x-1=0$

Now, this is a quadratic equation in 'x'. Here, $a=1,b=1,c=-1$

Applying quadratic formula to solve for 'x'.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x=\frac{-1\pm \sqrt{1^2-4(1)(-1)}}{2(1)}\\\\x=\frac{-1\pm \sqrt{1+4}}{2}\\\\x=\frac{-1\pm\sqrt5}{2}$

Replace 'x' by $\sin \theta$. This gives,

$\sin \theta=\frac{-1\pm \sqrt5}{2}$

Taking inverse of sine, we get:

$\theta=\sin^{-1}(\frac{-1\pm \sqrt5}{2})$