Question

How do you solve $ {\sin ^2}x + 2\cos x = 2 $ over the interval $ 0 < x < 2\pi $ ?

Answer 1

Hint: In order to determine the solution of the above trigonometric equation in the interval $ 0 < x < 2\pi $ replace the $ {\sin ^2}x $ using the identity of trigonometry $ {\sin ^2}x = 1 - {\cos ^2}x $ . Apply the formula $ {\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB $ by considering A as $ \cos x $ and B as 1. Derive the angle whose value of cosine is equal to 1 in the interval $ 0 < x < 2\pi $ to obtain the required answer.

Complete step by step solution:
We are given a trigonometric equation $ {\sin ^2}x + 2\cos x = 2 $ and we have to find the solution between $ 0 < x < 2\pi $
 $ {\sin ^2}x + 2\cos x = 2 $
Using the trigonometry $ {\sin ^2}x = 1 - {\cos ^2}x $ to replace $ {\sin ^2}x $ from the equation ,we get
 $ 1 - {\cos ^2}x + 2\cos x = 2 $
Simplifying it further we have
 $
\Rightarrow 1 - {\cos ^2}x + 2\cos x - 2 = 0 \\
\Rightarrow - {\cos ^2}x + 2\cos x - 1 = 0 \\
\Rightarrow {\cos ^2}x - 2\cos x + 1 = 0 \;
  $
Now applying the formula of square of difference of two numbers $ {\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB $ by considering A as $ \cos x $ and B as 1
 $
\Rightarrow {\left( {\cos x - 1} \right)^2} = 0 \\
\Rightarrow \cos x - 1 = 0 \\
\Rightarrow \cos x = 1 \;
  $
Taking inverse of cosine on both sides of the equation we get
 $ {\cos ^{ - 1}}\left( {\cos x} \right) = {\cos ^{ - 1}}\left( 1 \right) $
Since $ {\cos ^{ - 1}}\left( {\cos } \right) = 1 $ as they both are inverse of each other
 $ x = {\cos ^{ - 1}}\left( 1 \right) $
The value of x is an angle between $ 0 < x < 2\pi $ whose cosine value is equal to 1.
As we know $ \cos {0^ \circ } = 1 \to 0 = {\cos ^{ - 1}}\left( 1 \right) $
 $ x = 0 $
Therefore, the solution to the given trigonometric equation is \[x = 0\] between $ 0 < x < 2\pi $
So, the correct answer is “ \[x = 0\]”.

Note: 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] $ .