Question

How do you solve $2{\sin ^2}x = 5\sin x + 3$ ?

Answer 1

Hint: In this question, we have been asked to solve this quadratic equation. But the given equation is in terms of a trigonometric ratio. So, first, use substitution to replace this trigonometric ratio into an algebra. Then, convert the equation into the standard equation - $a{x^2} + bx + c = 0$. Find two numbers such that their product is equal to the coefficients of first and third term and their sum is equal to the middle term. Then, make factors and use them, find the value of the variable assumed. After that, find the value of $\sin x$.

Complete step by step solution:
Bring the right side of the equation to the left side.
Write the given equation in the quadratic form as follows
$2{\sin ^2}x - 5\sin x - 3 = 0$ ………..…. $\left( 1 \right)$
The above equation is of the form $a{x^2} + bx + c = 0$
Now, make substitution to the above quadratic equation as given below
Let $t = \sin x$
Equation $\left( 1 \right)$ can be written in the form as
$2{t^2} - 5t - 3 = 0$ ………...…. $\left( 2 \right)$
It can be factorized in the form as follows
$\Rightarrow$$2{t^2} - 6t + t - 3 = 0$
Taking $2t$ as common from the first two terms we get,
$\Rightarrow$$2t\left( {t - 3} \right) + 1\left( {t - 3} \right) = 0$
Making factors,
$\Rightarrow$$(2t + 1)\left( {t - 3} \right) = 0$
Now, we will keep each factor equal to 0,
$\Rightarrow$$(2t + 1) = 0,\,\left( {t - 3} \right) = 0$
Shifting to find the values,
$\Rightarrow$$2t = - 1,\,t = 3$
$\Rightarrow$$t = \dfrac{{ - 1}}{2},\,t = 3$
Now put $t = \dfrac{{ - 1}}{2},\,3$ in $t = \sin x$
$\dfrac{{ - 1}}{2} = \sin x,\,3 = \sin x$
$\sin x = 3$ is not possible, since it has a range of $\left[ { - 1,1} \right].$
Solving for $x$ in the equation, let us take the values of $\sin $ from the unit circle:
We get $\sin x = \dfrac{{ - 1}}{2}\,$ when $x$ is $\dfrac{{7\pi }}{6}$ or $\dfrac{{11\pi }}{6}$
These are the solutions, but it is not done completely.
We need to show that this value will stay the same after any full rotation. $x$ can also be $\dfrac{{ - \pi }}{6}$ because this is same as rotating $\dfrac{{11\pi }}{6}$
Add $2\pi k$ to both the solutions.
Here $k$ means any integer, $2\pi $ means a full rotation of the unit circle.
All together, it shows that the answer will be the same after any full rotation.

Hence the results are
$x = \dfrac{{7\pi }}{6} + 2\pi k,\,\dfrac{{11\pi }}{6} + 2\pi k$

Note: We can also solve this method using quadratic formula,
The given equation is of the quadratic form $a{x^2} + bx + c = 0$
The quadratic formula is
$t = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
From equation $\left( 2 \right)$
$a = 2,\,b = - 5,\,c = - 3$
Putting all the values,
$\Rightarrow$$t = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 2 \right)\left( { - 3} \right)} }}{{2\left( 2 \right)}}$
Simplifying the equation,
$\Rightarrow$$t = \dfrac{{\left( 5 \right) \pm \sqrt {\left( {25} \right) + 24} }}{4}$
$\Rightarrow$$t = \dfrac{{\left( 5 \right) \pm \sqrt {49} }}{4}$
$\Rightarrow$$t = \dfrac{{\left( 5 \right) \pm 7}}{4}$
Hence, we get,
$\Rightarrow$$t = \dfrac{{5 + 7}}{4},\dfrac{{5 - 7}}{4}$
$\Rightarrow$$t = \dfrac{{12}}{4},\dfrac{{ - 2}}{4}$
$\Rightarrow$$t = 3, - \dfrac{1}{2}$
Hence solving for $t$ after full rotation from the unit circle we get the required result as follows
$x = \dfrac{{7\pi }}{6} + 2\pi k,\,\dfrac{{11\pi }}{6} + 2\pi k$