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Question: How do you solve \(2{\sin ^2}x = 5\sin x + 3\) ?...

How do you solve 2sin⁡2x=5sin⁡x+32{\sin ^2}x = 5\sin x + 3 ?

Explanation

Solution

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 - ax2+bx+c=0a{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\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
2sin⁡2x−5sin⁡x−3=02{\sin ^2}x - 5\sin x - 3 = 0 ………..…. (1)\left( 1 \right)
The above equation is of the form ax2+bx+c=0a{x^2} + bx + c = 0
Now, make substitution to the above quadratic equation as given below
Let t=sin⁡xt = \sin x
Equation (1)\left( 1 \right) can be written in the form as
2t2−5t−3=02{t^2} - 5t - 3 = 0 ………...…. (2)\left( 2 \right)
It can be factorized in the form as follows
⇒\Rightarrow 2t2−6t+t−3=02{t^2} - 6t + t - 3 = 0
Taking 2t2t as common from the first two terms we get,
⇒\Rightarrow 2t(t−3)+1(t−3)=02t\left( {t - 3} \right) + 1\left( {t - 3} \right) = 0
Making factors,
⇒\Rightarrow (2t+1)(t−3)=0(2t + 1)\left( {t - 3} \right) = 0
Now, we will keep each factor equal to 0,
⇒\Rightarrow (2t+1)=0, (t−3)=0(2t + 1) = 0,\,\left( {t - 3} \right) = 0
Shifting to find the values,
⇒\Rightarrow 2t=−1, t=32t = - 1,\,t = 3
⇒\Rightarrow t=−12, t=3t = \dfrac{{ - 1}}{2},\,t = 3
Now put t=−12, 3t = \dfrac{{ - 1}}{2},\,3 in t=sin⁡xt = \sin x
−12=sin⁡x, 3=sin⁡x\dfrac{{ - 1}}{2} = \sin x,\,3 = \sin x
sin⁡x=3\sin x = 3 is not possible, since it has a range of [−1,1].\left[ { - 1,1} \right].
Solving for xx in the equation, let us take the values of sin⁡\sin from the unit circle:
We get sin⁡x=−12 \sin x = \dfrac{{ - 1}}{2}\, when xx is 7π6\dfrac{{7\pi }}{6} or 11π6\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. xx can also be −π6\dfrac{{ - \pi }}{6} because this is same as rotating 11π6\dfrac{{11\pi }}{6}
Add 2πk2\pi k to both the solutions.
Here kk means any integer, 2π2\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=7π6+2πk, 11π6+2πkx = \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 ax2+bx+c=0a{x^2} + bx + c = 0
The quadratic formula is
t=−b±b2−4ac2at = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
From equation (2)\left( 2 \right)
a=2, b=−5, c=−3a = 2,\,b = - 5,\,c = - 3
Putting all the values,
⇒\Rightarrow t=−(−5)±(−5)2−4(2)(−3)2(2)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=(5)±(25)+244t = \dfrac{{\left( 5 \right) \pm \sqrt {\left( {25} \right) + 24} }}{4}
⇒\Rightarrow t=(5)±494t = \dfrac{{\left( 5 \right) \pm \sqrt {49} }}{4}
⇒\Rightarrow t=(5)±74t = \dfrac{{\left( 5 \right) \pm 7}}{4}
Hence, we get,
⇒\Rightarrow t=5+74,5−74t = \dfrac{{5 + 7}}{4},\dfrac{{5 - 7}}{4}
⇒\Rightarrow t=124,−24t = \dfrac{{12}}{4},\dfrac{{ - 2}}{4}
⇒\Rightarrow t=3,−12t = 3, - \dfrac{1}{2}
Hence solving for tt after full rotation from the unit circle we get the required result as follows
x=7π6+2πk, 11π6+2πkx = \dfrac{{7\pi }}{6} + 2\pi k,\,\dfrac{{11\pi }}{6} + 2\pi k