Question

How do you solve $ {\sin ^2}\theta + \cos \theta = 2 $ ?

Answer 1

Hint: In order to determine the solution of the above trigonometric equation replace the $ \sin x $ as $ t $ . Compare the given quadratic equation with the standard form $ a{x^2} + bx + c $ to obtain the values for the variables. Now find the value of determinant using formula $ D = {b^2} - 4ac $ . You will get $ D < 0 $ so the roots are imaginary. So no solution exists for the given equation.

Complete step by step solution:
We are given a trigonometric equation $ {\sin ^2}\theta + \cos \theta = 2 $ and we have to find its solution
 $ {\sin ^2}\theta + \cos \theta = 2 $
Using the trigonometry identity $ {\sin ^2}x = 1 - {\cos ^2}x $ to replace $ {\sin ^2}x $ from the equation ,we get
 $ 1 - {\cos ^2}\theta + \cos \theta = 2 $
Rearranging the terms in the standard quadratic form $ a{x^2} + bx + c $
 $ {\cos ^2}\theta - \cos \theta + 1 = 0 $
Lets $ t = \cos x $ . So substituting $ \sin x\,as\,t $ in the equation, we get
 $ {t^2} - t + 1 = 0 $
We have obtained a quadratic equation in $ t $ . Comparing the above quadratic equation with the standard quadratic equation $ a{x^2} + bx + c $ , we have
a=1
b=-1
c=1
Let’s find the determinant of the above quadratic equation to understand the nature of roots by using the formula
 $ D = {b^2} - 4ac $
Putting the values of the variable, we get
 $
\Rightarrow D = {\left( { - 1} \right)^2} - 4\left( 1 \right)\left( 1 \right) \\
   = 1 - 4 \\
   = - 3 \;
  $
Since, $ D < 0 $ then both the roots of the quadratic equation are complex . SO there is no real root.
Therefore, there exists no solution for the given trigonometric equation.
So, the correct answer is “there exists no solution ”.

Note: 1.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 .
The degree of the quadratic equation is of the order 2.
Even Function – A function $ f(x) $ is said to be an even function ,if $ f( - x) = f(x) $ for all x in its domain.
Odd Function – A function $ f(x) $ is said to be an even function ,if $ f( - x) = - f(x) $ for all x in its domain.
Period of cosine function is $ 2\pi $ .
The domain of cosine function is in the interval $ \left[ {0,\pi } \right] $ and the range is in the interval $ \left[ { - 1,1} \right] $ .