Question

Help please
Solve the given equation.
sin2 θ − 5 sin θ − 6 = 0

Answer 1

Do you mean\[2\sin\theta-5\sin\theta-6=0\]?

Answer 2

yes

Answer 3

If that's the case, then you can just add the sine terms and solve for theta
\[2 \sin \theta - 5 \sin \theta - 6=0\]
\[-3 \sin \theta - 6 =0\]

If the expression is

\[ \sin ^2 \theta - 5\sin \theta - 6 = 0\]
you can treat "sin theta" like an x in the normal polynomial

\[x^2-5x-6=0\]
and use the quadratic formula to find two values for sin theta

\[ \sin\theta = \frac{-b±\sqrt{b^2-4ac}}{2a} = \frac{-(-5)±\sqrt{(-5)^2-4(1)(-6)}}{2(1)}\]