How do you solve $ {\cos ^2}(x) + sinx = 1 $ ?

Hint: We can simplify the given question by using the trigonometric ratios or identities and convert all the terms in the form of one trigonometric ratio so that on replacing the trigonometric ratio in the given equation with any other variable, the equation is a polynomial equation. In a polynomial equation, there exist exactly as many roots as its degree, so on converting the given equation in polynomial form, we can find out its degree and then we can find the solutions by factoring the equation or by a special formula called completing the square method. So, after finding the possible values of the trigonometric ratio, we can find the values of x.

Complete step-by-step answer:
We know that,
 $
  {\sin ^2}x + {\cos ^2}x = 1 \\
   \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x \;
  $
Using this value of $ {\cos ^2}x $ in the given question, we get –
 $
  {\cos ^2}x + \sin x = 1 \\
   \Rightarrow 1 - {\sin ^2}x + \sin x = 1 \\
   \Rightarrow {\sin ^2}x - \sin x = 0 \;
  $
Now, the equation obtained is quadratic. The degree of this equation is 2, so its number of solutions is 2.
Taking $ \sin x $ common, we get –
 $
  \sin x(\sin x - 1) = 0 \\
   \Rightarrow \sin x = 0,\,\sin x - 1 = 0 \\
   \Rightarrow \sin x = 0,\,\sin x = 1 \;
  $
When,
 $
  \sin x = 0 \\
   \Rightarrow x = 0,\pi ,2\pi ,3\pi ... \;
  $
And
 $
  \sin x = 1 \\
   \Rightarrow x = \dfrac{\pi }{2},\dfrac{{5\pi }}{2}... \;
  $
Now under the limit $ [0,2\pi ] $ the possible values of x are $ 0,\dfrac{\pi }{2},\pi ,2\pi $
Hence, when $ {\cos ^2}x + \sin x = 1,\,x = 0,\dfrac{\pi }{2},\pi ,2\pi $ .
So, the correct answer is “ $ {\cos ^2}x + \sin x = 1,\,x = 0,\dfrac{\pi }{2},\pi ,2\pi $ ”.

Note: All the trigonometric functions are interrelated with each other and one can be converted into another using this knowledge or the trigonometric identities like we have used the identity $ {\sin ^2}x + {\cos ^2}x = 1 $ in the given question. $ a{x^2} + bx + c = 0 $ is the standard form of a quadratic equation and equations of this form are solved using factorization. In the polynomial obtained, there is no constant c, so it becomes easier to solve the equation, that is the equation can be solved by just taking out x common.