Question

How do you solve ${\sin ^2}x - 8\sin x - 4 = 0$ and find all values of x in the interval $\left[ {0,{{360}^0}} \right)$ ?

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 distinct and real. Obtain the roots using quadratic formula and put the value of $t$ in the roots to obtain the required solution.

Complete step by step solution:
We are given a trigonometric equation ${\sin ^2}x - 8\sin x - 4 = 0$ and we have to find the solution in the interval $\left[ {0,{{360}^0}} \right)$
Let $t = \sin x$ . So substituting $\sin x\,as\,t$ in the equation, we get
${t^2} - 8t - 4 = 0$
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=-8
c=-4
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
$\Rightarrow D = {\left( { - 8} \right)^2} - 4\left( 1 \right)\left( { - 4} \right) \\\ = 64 + 16 \\\ = 80 \;$
Since, $D > 0$ then both the roots of the quadratic equation are distinct and real.
Root of the equation will be
$\Rightarrow {t_1} = \dfrac{{ - b}}{{2a}} + \dfrac{{\sqrt D }}{{2a}} = \dfrac{{ - \left( { - 8} \right)}}{2} + \dfrac{{\sqrt {80} }}{2} = 4 + \dfrac{{4\sqrt 5 }}{2} = 4 + 2\sqrt 5 \approx 8.472 \\\ \Rightarrow {t_2} = \dfrac{{ - b}}{{2a}} - \dfrac{{\sqrt D }}{{2a}} = \dfrac{{ - \left( { - 8} \right)}}{2} - \dfrac{{\sqrt {80} }}{2} = 4 - \dfrac{{4\sqrt 5 }}{2} = 4 - 2\sqrt 5 \approx - 0.47 \;$
Value of ${t_1}$ is rejected as the range of sine function is in the interval $\left[ { - 1,1} \right]$ .
So we are left with
${t_2} = - 0.47$
Putting back the value of $t$
$\Rightarrow \sin x = - 0.47 \\\ \Rightarrow x = {\sin ^{ - 1}}\left( { - 0.47} \right) \;$
The value of x is an angle having sine value equal to $- 0.47$ . Since the value $- 0.47$ is not a remarkable value. Use the calculator to find the answer.
Sine function is negative in the 3rd and 4th quadrant . so we have two solution for the value of x
For the third quadrant we have
$\Rightarrow x = \pi + {\sin ^{ - 1}}\left( {0.47} \right) \\\ \Rightarrow x = 180 + 28.16 \\\ \Rightarrow x = {208.16^ \circ } \;$
And for the fourth quadrant, we have
$\Rightarrow x = 2\pi - {\sin ^{ - 1}}\left( {0.47} \right) \\\ \Rightarrow x = 360 - 28.16 \\\ \Rightarrow x = {331.84^ \circ } \;$
Therefore, the solution of the given trigonometric equation is $x = {208.16^ \circ },{331.84^ \circ }$
So, the correct answer is “ $x = {208.16^ \circ },{331.84^ \circ }$”.

Note : 1. The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
If D is equal to zero, then both of the roots will be the same and real.
If D is a positive number then, both of the roots are real solutions.
If D is a negative number, then the root are the pair of complex solutions
Also remember,
1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of sine function is $2\pi$ .
3. The domain of sine function is in the interval $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$ and the range is in the interval $\left[ { - 1,1} \right]$ .
4.Don’t forget to rearrange quadratic equations in the standard form.
5. Don’t forget to compare the given quadratic equation with the standard one every time.