Question

How do you solve the trigonometric equation $2{\left( {\sin x} \right)^2} - \cos x - 1 = 0$ to find the values of x in the interval $\left[ {0,2\pi } \right]$ that satisfy the given equation.

Answer 1

First convert ${\sin ^2}x$ into ${\cos ^2}x$ by the use of trigonometric identity then assume $\cos x$ as $y$ and obtain the quadratic equation. Solve the quadratic equation by the use of middle term splitting method and thus get the values of y. After that replace y as $\cos x$ and then get the values of x.

Complete step by step solution:
We are given that we are required to solve $2{\left( {\sin x} \right)^2} - \cos x - 1 = 0$.
So, first convert the sine term into cosine by the use of trigonometric formula ${\sin ^2}x = 1 - {\cos ^2}x$ as shown below.
2\left\\{ {1 - {{\left( {\cos x} \right)}^2}} \right\\} - \cos x - 1 = 0
Simplify the equation as shown below.
$\Rightarrow 2 - 2{\left( {\cos x} \right)^2} - \cos x - 1 = 0$
$\Rightarrow - 2{\left( {\cos x} \right)^2} - \cos x + 1 = 0$
Multiply both sides of the equation by negative one and write the result.
$\Rightarrow 2{\left( {\cos x} \right)^2} + \cos x - 1 = 0$
Let us assume that y = cos x.
Replacing this in the given equation, we will obtain the following equation: -
$\Rightarrow 2{y^2} + y - 1 = 0$
Now, we can write the equation in the above line as following also: -
$\Rightarrow 2{y^2} + 2y - y - 1 = 0$
Now, taking 2y common from the first two terms in the left hand side of above expression, we will then obtain the following expression: -
$\Rightarrow 2y\left( {y + 1} \right) - y - 1 = 0$
Now, taking $- 1$ common from the last two terms in the left hand side of the above expression, we will then obtain the following expression: -
$\Rightarrow 2y\left( {y + 1} \right) - 1\left( {y + 1} \right) = 0$
Now taking the factor (y+1) common from the above equation, we will then obtain the following expression: -
$\Rightarrow \left( {y + 1} \right)\left( {2y - 1} \right) = 0$
Now, we have got two values of y which are $\dfrac{1}{2}$ and $- 1$.
Putting back y as cos x as we assumed in the beginning, we will then obtain that the two possible values of cos (x) are $\dfrac{1}{2}$ and $- 1$. But since we know that sine of any function always lies between – 1 and 1 also the values of x in the interval $\left[ {0,2\pi } \right]$ that satisfy the given equation.
Therefore, values of cos(x) that satisfy the equation $\cos \left( x \right) = \dfrac{1}{2}$ and in the interval $\left[ {0,2\pi } \right]$ are $\dfrac{\pi }{6}$ and $\dfrac{{5\pi }}{3}$.
Similarly, values of cos(x) that satisfy the equation $\cos \left( x \right) = - 1$ and in the interval $\left[ {0,2\pi } \right]$ is $\pi$ only.
Thus with in the interval $\left[ {0,2\pi } \right]$, the solutions for x are $\dfrac{\pi }{6}$, $\pi$ and $\dfrac{{5\pi }}{3}$.

Note:
The students must note that in the given solution we used splitting of middle term method but you may also use the fact that the roots of $a{y^2} + by + c = 0$ are given by:-
$\Rightarrow y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$