Question

How do you solve $2{(\sin x)^2} - 5\cos x - 4 = 0$ in the interval $[0,2\pi ]$ ?

Answer 1

Write the equation in terms of $\cos x$ , Convert it into a quadratic expression which is an equation of degree $2$ , and then solve it. After getting the solution of the quadratic expression, find the range of values of $\cos x$ in the interval $[0,2\pi ]$ .

Formula used: The roots of a quadratic equation( $a{x^2} + bx + c = 0$ ) can be found by the formula,
$x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$

Complete step-by-step solution:
The given expression, $2{(\sin x)^2} - 5\cos x - 4 = 0$
From the formula, ${\sin ^2}x + {\cos ^2}x = 1$ We can write ${\sin ^2}x = 1 - {\cos ^2}x$
$\Rightarrow 2[1 - {\cos ^2}x] - 5\cos x - 4 = 0$
Multiplying 2 with the contents inside the bracket i.e. $1 - {\cos ^2}x$
$\Rightarrow [2 - 2{\cos ^2}x] - 5\cos x - 4 = 0$
After further evaluating the constants,
$\Rightarrow - 2{\cos ^2}x - 5\cos x - 2 = 0$
Multiplying the entire equation with ( $-$ ) on both sides,
$\Rightarrow 2{\cos ^2}x + 5\cos x + 2 = 0$
Now, Let’s assume $x = \cos x$ . By this assumption, we get a Polynomial equation of degree $2$ (quadratic).
$\Rightarrow 2{x^2} + 5x + 2 = 0$
The roots of a quadratic equation( $a{x^2} + bx + c = 0$ ) can be found by the formula,
$x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$
In our solution, $a = 2;b = 5;c = 2$ .On substituting in the Quadratic formula,
$\Rightarrow x = \left[ {\dfrac{{ - 5 \pm \sqrt {{5^2} - 4 \times 2 \times 2} }}{{2 \times 2}}} \right]$
Evaluate the expression inside the root first.
$\Rightarrow x = \left[ {\dfrac{{ - 5 \pm \sqrt {25 - 16} }}{{2 \times 2}}} \right]$
$\Rightarrow x = \left[ {\dfrac{{ - 5 \pm \sqrt 9 }}{{2 \times 2}}} \right]$
Now evaluate the square root of $9$
$\Rightarrow x = \left[ {\dfrac{{ - 5 \pm 3}}{{2 \times 2}}} \right]$
On simplifying we get,
$\Rightarrow x = \left[ {\dfrac{{ - 5 \pm 3}}{4}} \right]$
This comes to two cases,
$\Rightarrow x = \left[ {\dfrac{{ - 5 + 3}}{4}} \right];x = \left[ {\dfrac{{ - 5 - 3}}{4}} \right]$
$\Rightarrow x = \left[ {\dfrac{{ - 2}}{4}} \right];x = \left[ {\dfrac{{ - 8}}{4}} \right]$
Evaluate further to get simple constants.
$\Rightarrow x = \dfrac{{ - 1}}{2};x = - 2$
Now coming back to the step where we assumed $x = \cos x$ , Substitute back $x = \cos x$ in the solutions.
$\Rightarrow \cos x = \dfrac{{ - 1}}{2};\cos x = - 2$
Checking if the values of $\cos x$ exist in the interval $[0,2\pi ]$ ,
The expression, $\cos x = - 2$
$\cos x$ cannot be equal to $- 2$ Hence for no value of $x$ the above condition can be satisfied. The maximum value of $\cos x$ is $1$ and minimum value of $\cos x$ is $- 1$ . The range of $\cos x = \\{ - 1,1\\}$ , so there cannot be a value of $x$ for which $\cos x = - 2$ .
The expression, $\cos x = \dfrac{{ - 1}}{2}$ Since $\cos x$ is negative in ${2^{nd}}$ and ${3^{rd}}$ Quadrant,
So, the values of $x$ should lie in ${2^{nd}},{3^{rd}}$ Quadrant in the interval $[0,2\pi ]$ . There are two values of $\cos x$ in this range which are,
$\Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$
$\Rightarrow x={{\cos }^{-1}}\left( \pi \pm \dfrac{\pi }{3} \right)$

Therefore the solution for the expression $2{(\sin x)^2} - 5\cos x - 4 = 0$ in the interval $[0,2\pi ]$ is $x = \dfrac{{2\pi }}{3};\dfrac{{4\pi }}{3}$

Note: Must check where the Trigonometric functions become negative in which Quadrant to easily find the values in the given range. Convert the entire Trigonometric equation in either $\cos x$ or $\sin x$ to easily solve the entire expression as a whole.