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 $\left[ { - 1,1} \right]$ .