Question

How do you solve \[{\sin ^2}x = 2\cos x + 2\] from 0 to $2\pi $?

Answer 1

Hint:
According to the question we have to solve the given trigonometric expression which is \[{\sin ^2}x = 2\cos x + 2\]from 0 to$2\pi $. So, first of all to determine the solution of the given trigonometric expression from 0 to$2\pi $, we have to use the formula or identity which is as mentioned below:
Formula used:
$ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1................(A)$
Now, we have to rearrange the terms of the identity (A) as mentioned just above, to obtain an expression in the form of $\cos x$.
Now, we have to rearrange the terms of the given trigonometric expression which is\[{\sin ^2}x = 2\cos x + 2\] which can be done with the help of the identity or formula as mentioned below:
Formula used:
$ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x.......................(B)$
Now, after rearranging the terms of the given trigonometric expression we have to let $\cos x$ any variable.
Now, we will obtain an quadratic expression so, we have to obtain the roots/zeros of the quadratic expression which can be done by obtaining the coefficient of u (as we let) with the help of the coefficient of ${u^2}$and the constant term.
Now, on solving the quadratic expression we will obtain two roots/zeros and on solving them we can easily determine the values of x.
Now, we have to replace the value of u as we let to obtain the required solution for the given expression.
Formula used:
$
   \Rightarrow \cos {0^{\circ}} = 1 \\
   \Rightarrow \cos 2\pi = 1 \\
 $
$ \Rightarrow {\cos ^{ - 1}}(\cos x) = x$

Complete step by step solution:
Step 1: First of all to determine the solution of the given trigonometric expression from 0 to $2\pi $, we have to use the formula or identity (A), which is as mentioned in the solution hint. Hence,
$ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
Now, we have to rearrange the terms of the identity (A) as mentioned just above, to obtain an expression in the form of $\cos x$.
$ \Rightarrow {\sin ^2}x + 2\sin x = 2$
Step 2: Now, we have to rearrange the terms of the given trigonometric expression which is\[{\sin ^2}x = 2\cos x + 2\]which can be done with the help of the identity or formula (B) as mentioned in the solution hint. Hence,
\[ \Rightarrow 1 - {\cos ^2}x + 2\cos x = 2\]
On multiplying with negative sign in the both sides of the expression obtained just above,
\[ \Rightarrow {\cos ^2}x - 2\cos x + 2 = 0\]
Step 3: Now, after rearranging the terms of the given trigonometric expression we have to let $\cos x$ variable u. Hence, Let,
$ \Rightarrow \cos x = u$
Step 4: Now, we will obtain an quadratic expression so, we have to obtain the roots/zeroes of the quadratic expression which can be done by obtaining the coefficient of u (as we let) with the help of the coefficient of ${u^2}$and the constant term. Hence,
$
   \Rightarrow {u^2} - 2u + 1 = 0 \\
   \Rightarrow {u^2} - (1 + 1)u + 1 = 0 \\
   \Rightarrow {u^2} - u - u + 1 = 0 \\
 $
On taking the terms common which can be,
$
   \Rightarrow u(u - 1) - 1(u - 1) = 0 \\
   \Rightarrow (u - 1)(u - 1) = 0 \\
 $
Step 5: Now, on solving the quadratic expression we will obtain two roots/zeroes and on solving them we can easily determine the values of x. Hence,
$ \Rightarrow u = 1$
Step 6: Now, we have to replace the value of u as we let to obtain the required solution for the given expression. Hence,
$ \Rightarrow \cos x = 1$
Now, as we know that,
$ \Rightarrow x = {\cos ^{ - 1}}(1)$
Step 7: Now, to solve the expression as obtained in the solution step 6 we have to use the formula (C) which is as mentioned in the solution hint. Hence,
$
   \Rightarrow x = {\cos ^{ - 1}}(\cos {0^{\circ}}) \\
   \Rightarrow x = 0 \\
 $
And,
$
   \Rightarrow x = {\cos ^{ - 1}}(\cos 2\pi ) \\
   \Rightarrow x = 2\pi \\
 $

Hence, with the help of formulas (A), (B), and (C) we have determined the required solution which is $x = 0,2\pi $

Note:
To obtain the solution of the given trigonometric expression or we can say that to obtain the roots/zeroes it is necessary that we have to convert $\sin x$to $\cos x$with the help identity which is mentioned in the solution hint.
To obtain the factors and roots of the obtained quadratic expression which is in trigonometric form to solve it easily we can let the obtained trigonometric term to some variable.