How do you solve $ {\sin ^2}A = \cos A - 1 $ ?

Hint: In order to determine the solution of the above trigonometric equation replace the $ {\sin ^2}x $ using the identity of trigonometry $ {\sin ^2}x = 1 - {\cos ^2}x $ . We will obtain a quadratic equation. Factorise the quadratic equation using splitting by the middle term and then equate every factor equal to zero to obtain the required solution.

Complete step by step solution:
We are given a trigonometric equation $ {\sin ^2}A = \cos A - 1 $ and we have to find the solution
 $ {\sin ^2}A = \cos A - 1 $
Using the trigonometry $ {\sin ^2}x = 1 - {\cos ^2}x $ to replace $ {\sin ^2}x $ from the equation ,we get
 $ 1 - {\cos ^2}A = \cos A - 1 $
Rearranging the terms in the standard quadratic form $ a{x^2} + bx + c $
 $ {\cos ^2}A + \cos A - 2 = 0 $
Factoring the above quadratic equation by using the middle term splitting method by splitting $ \cos A\,as\,2\cos A - \cos A $ . Our equation now becomes
 $ {\cos ^2}A + 2\cos A - \cos A - 2 = 0 $
Pulling out common from first two terms and last two terms
 $ \cos A\left( {\cos A + 2} \right) - 1\left( {\cos A + 2} \right) = 0 $
Finding the common binomial parenthesis, the equation becomes
 $ \left( {\cos A + 2} \right)\left( {\cos A - 1} \right) = 0 $ ----(1)
Dividing both sides of equation with $ \left( {\cos A - 1} \right) $ , we have
 \[
  \dfrac{1}{{\left( {\cos A - 1} \right)}} \times \left( {\cos A + 2} \right)\left( {\cos A - 1} \right) = 0 \times \dfrac{1}{{\left( {\cos A - 1} \right)}} \\
  \left( {\cos A + 2} \right) = 0 \\
  \cos A = - 2 \;
 \]
Since the range of cosine function is in the interval $ \left[ { - 1,1} \right] $ so the above is not possible
Now dividing the equation (1) with \[\left( {\cos A + 2} \right)\] , we get
 \[
  \dfrac{1}{{\left( {\cos A + 2} \right)}} \times \left( {\cos A + 2} \right)\left( {\cos A - 1} \right) = 0 \times \dfrac{1}{{\left( {\cos A + 2} \right)}} \\
  \left( {\cos A - 1} \right) = 0 \\
\Rightarrow \cos A = 1 \\
  A = {\cos ^{ - 1}}\left( 1 \right) \;
 \]
The value of A is an angle having cosine value 1. Since we know $ \cos 0 = 1 \to 0 = {\cos ^{ - 1}}\left( 1 \right) $ .
And we know the period of cosine function is $ 2\pi $ as it repeats itself after every $ 2\pi $ interval. We have value of A as
 \[A = 2n\pi \] where n is any integer value
Therefore the solution of the given trigonometric equation is \[A = 2n\pi \] , n is any integer.
So, the correct answer is “ \[A = 2n\pi \] ”.

Note: Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $ a{x^2} + bx + c $ where $ x $ is the unknown variable and a,b,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become linear equation and will no more quadratic .
1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of cosine function is $ 2\pi $ .
3. The domain of cosine function is in the interval $ \left[ {0,\pi } \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .
2.Don’t forget to rearrange quadratic equations in the standard form.
3. Write the factors when the middle term is split using the proper sign.