How do you solve \[3{\cos ^2}x + 5\sin x - 1 = 0\] ?

Hint: The given question has two trigonometric functions involved. We will first convert them into one trigonometric function. We will turn the equation in sin form. That will then be a quadratic equation. After that we will solve it using the factorization method. After that we will find the value of x using the value of sin x.

Complete step by step answer:
Given the equation is, \[3{\cos ^2}x + 5\sin x - 1 = 0\]
Now we know that, \[{\cos ^2}x = 1 - {\sin ^2}x\]
So replacing it we get,
\[3\left( {1 - {{\sin }^2}x} \right) + 5\sin x - 1 = 0\]
On multiplying the brackets,
\[3 - 3{\sin ^2}x + 5\sin x - 1 = 0\]
This can be simplified as,
\[ - 3{\sin ^2}x + 5\sin x + 2 = 0\]
Multiplying both sides by the minus sign,
\[3{\sin ^2}x - 5\sin x - 2 = 0\]
\[3{\left( {\sin x} \right)^2} - 5\sin x - 2 = 0\]
Replacing sin by u,
\[3{u^2} - 5u - 2 = 0\]
This is a quadratic equation of the form \[a{x^2} + bx + c = 0\]
This equation is solved by using the factorisation method,
\[3{u^2} - 6u + u - 2 = 0\]
Taking 3u common from first two terms and 1 common from last two terms,
\[3u\left( {u - 2} \right) + 1\left( {u - 2} \right) = 0\]
\[\left( {u - 2} \right)\left( {3u + 1} \right) = 0\]
Now separately equating the brackets,
\[u - 2 = 0\] or \[3u + 1 = 0\]
Re-substituting the value of u,
\[\sin x = 2\] or \[\sin x = \dfrac{{ - 1}}{3}\]
But as we know \[\sin x = 2\] is not the perfect value as the sine function range is from $-1$ to $1$.
Therefore, the required solution is \[\sin x = \dfrac{{ - 1}}{3}\].

Note:
This is a simple question to solve. All we need to observe is the hidden quadratic equation in trigonometric functions. We used the common factors method to solve the quadratic equation. Note that we can also use the quadratic formula to find the roots. After getting the solutions from the quadratic equation, we have to make sure about the range of trigonometric functions.