Question

How do you solve $- 3\cos x + 3 = 2{\sin ^2}x$ in the interval $0 \leqslant x \leqslant 2\pi$?

Answer 1

Hint : In order to solve the given equation first convert $\sin x$ in terms of $\cos x$. Then by the use of the mid-term factorization method or Quadratic equation method we get two roots. Comparing the roots with trigonometric values we get our respective angles. Then look for the range given and according to that find the values.

Complete step by step solution:
We are given with the equation $- 3\cos x + 3 = 2{\sin ^2}x$. Write the equation in equation in one side and we get:
$2{\sin ^2}x + 3\cos x - 3 = 0$
Convert $\sin x$ in terms of $\cos x$ and this can be done by the formula we know that is ${\sin ^2}x = 1 - {\cos ^2}x$
Put this value in the equation and we get:
$2\left( {1 - {{\cos }^2}x} \right) + 3\cos x - 3 = 0$.
Solving it further we get:
$2\left( {1 - {{\cos }^2}x} \right) + 3\cos x - 3 = 0 \\\ 2 - 2{\cos ^2}x + 3\cos x - 3 = 0 \\\ \- 2{\cos ^2}x + 3\cos x - 1 = 0 \\\ 2{\cos ^2}x - 3\cos x + 1 = 0 \;$
We obtained a Quadratic equation:
Let’s solve this equation with the Quadratic formula:
Let’s first compare the given equation $2{\cos ^2}x - 3\cos x + 1 = 0$ with the standard quadratic equation $a{x^2} + bx + c$ to get the values of $a,b,c$ where $x = \cos x$, we get
$a = 2 \\\ b = - 3 \\\ c = 1 \;$
Determinant $D$ of quadratic equation is given as $D = {b^2} - 4ac$
Putting the values of $a,b,c$, we get the determinant as

$$D = {\left( { - 3} \right)^2} - 4\left( 2 \right)\left( 1 \right) \\\ D = 9 - 8 \\\ D = 1 \;$$

Since, we got $D > 0$, which means there are two distinct real roots or in other words the equation has two x-intercepts.
The Root or intercepts of x-axis are $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
$\Rightarrow {x_1} = \dfrac{{ - \left( { - 3} \right) + \sqrt 1 }}{{2\left( 2 \right)}} \\\ {x_1} = \dfrac{{3 + 1}}{4} = \dfrac{4}{4} \\\ {x_1} = 1 \\\ \Rightarrow {x_2} = \dfrac{{ - \left( { - 3} \right) - \sqrt 1 }}{{2\left( 2 \right)}} \\\ {x_2} = \dfrac{{3 - 1}}{4} = \dfrac{2}{4} \\\ {x_2} = \dfrac{1}{2} \\\$
Since, we had taken $x = \cos x$. So:
$\cos x = 1$ and $\cos x = \dfrac{1}{2}$
We know that $\cos 0 = 1$ So comparing it with first root we get:

$$\cos x = 1 \\\ \cos 0 = 1 \\\ \cos x = \cos 0 \\\ x = 0 \;$$

Similarly, we know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ So comparing it with second root we get:

$$\cos x = \dfrac{1}{2} \\\ \cos \dfrac{\pi }{3} = \dfrac{1}{2} \\\ \cos x = \cos \dfrac{\pi }{3} \\\ x = \dfrac{\pi }{3} \;$$

We are given a range of $0 \leqslant x \leqslant 2\pi$. And according to the rules we know that $\cos x$ is positive in 1st and 4th Quadrant.
We have the 1st quadrant value. Let’s check for 4th Quadrant that is:
$\cos \dfrac{\pi }{3} = \cos \left( {2\pi - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{{5\pi }}{3}} \right)$
Therefore, the values obtained from the equation $- 3\cos x + 3 = 2{\sin ^2}x$ in the interval $0 \leqslant x \leqslant 2\pi$ is:
$x = 0$ , $x = \dfrac{\pi }{3}$ and $x = \dfrac{{5\pi }}{3}$ .
So, the correct answer is “ $x = 0$ , $x = \dfrac{\pi }{3}$ and $x = \dfrac{{5\pi }}{3}$ .”.

Note : Always check the range before coming to the conclusion.
Conversion of the equation into the same variable value is important, if not changed may lead to an error. Look for the homes of trigonometric values, which are their quadrants otherwise only half value would be obtained.