Question

Use trigonometric identities, algebraic methods, and inverse trigonometric functions, as necessary, to solve the following trigonometric equation on the interval [0, 21).Round your answer to four decimal places, if necessary. If there is no solution, indicate "No Solution."3sin?(x) + 13sin(x) = -12

Answer 1

Answer:

No solution

Explanation:

The equation

$3(\sin x)^2+13\sin x=-12$

looks very much like a quadratic equation. Therefore, for the moment we say that

$y=\sin x$

and write the above equation as

$\begin{gathered} 3y^2+13y=-12 \\ \Rightarrow3y^2+13y+12=0 \end{gathered}$

Using the quadratic formula, we find that the solutions to the above equation are given by

$y=\frac{-13\pm\sqrt[]{13^2-4(3)(12)}}{2\cdot3}$$\begin{gathered} y=-\frac{4}{3} \\ y=-3 \end{gathered}$

Reminding ourselves that actually y was sin(x) gives

$undefined$