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Question: How do you solve \[2{\sin ^2}\theta + 3\sin \theta + 1 = 0\] from \[\left[ {0,2pi} \right]\]?...

How do you solve 2sin⁡2θ+3sin⁡θ+1=02{\sin ^2}\theta + 3\sin \theta + 1 = 0 from [0,2pi]\left[ {0,2pi} \right]?

Explanation

Solution

Hint : Here in this question, we have to find the value of θ\theta , the given equation is in the form of a quadratic equation. This is a quadratic equation for the variable sin⁡θ\sin \theta . By using the formula sin⁡θ=−b±b2−4ac2a\sin \theta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}, we can determine the roots and hence find the value of θ\theta .

Complete step-by-step answer :
The question involves the quadratic equation. To the quadratic equation we can find the roots by factorising or by using the formula sin⁡θ=−b±b2−4ac2a\sin \theta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}. So the equation is written as 2sin⁡2θ+3sin⁡θ+1=02{\sin ^2}\theta + 3\sin \theta + 1 = 0.
In general, the quadratic equation is represented as ax2+bx+c=0a{x^2} + bx + c = 0, when we compare the above equation to the general form of equation the values are as follows. a=2 b=3 and c=1. Now substituting these values to the formula for obtaining the roots we have
sin⁡θ=−3±32−4(2)(1)2(2)\sin \theta = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4(2)(1)} }}{{2(2)}}
On simplifying the terms, we have
⇒sin⁡θ=−3±9−84\Rightarrow \sin \theta = \dfrac{{ - 3 \pm \sqrt {9 - 8} }}{4}
Now subtract 8 from 9 we get
⇒sin⁡θ=−3±14\Rightarrow \sin \theta = \dfrac{{ - 3 \pm \sqrt 1 }}{4}
The number 1 is a perfect square so we can take out from square root we have
⇒sin⁡θ=−3±14\Rightarrow \sin \theta = \dfrac{{ - 3 \pm 1}}{4}
Therefore, we have sin⁡θ1=−3+14=−24=−12\sin {\theta _1} = \dfrac{{ - 3 + 1}}{4} = - \dfrac{2}{4} = - \dfrac{1}{2} or sin⁡θ2=−3−14=−44=−1\sin {\theta _2} = \dfrac{{ - 3 - 1}}{4} = \dfrac{{ - 4}}{4} = - 1
The value of θ\theta can be determined by
θ1=sin⁡−1(−12){\theta _1} = {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) and θ2=sin⁡−1(−1){\theta _2} = {\sin ^{ - 1}}( - 1)
So we have a table for the trigonometry ratio sine for the standard angles.

Angle030456090
sine012\dfrac{1}{2}12\dfrac{1}{{\sqrt 2 }}32\dfrac{{\sqrt 3 }}{2}1

By the ASTC rule the sine is negative in the third and fourth quadrant.
By the table for standard angles, we get
θ1=−π6{\theta _1} = - \dfrac{\pi }{6} and θ2=−π2{\theta _2} = - \dfrac{\pi }{2}
Therefore by the applying ASTC rule and table of trigonometry ratios we have
θ=5π6\theta = \dfrac{{5\pi }}{6}or 11π6\dfrac{{11\pi }}{6}or 3π2\dfrac{{3\pi }}{2}
Hence, we have found the value of θ\theta
So, the correct answer is “θ=5π6\theta = \dfrac{{5\pi }}{6}or 11π6\dfrac{{11\pi }}{6}or 3π2\dfrac{{3\pi }}{2}”.

Note : The quadratic equation can be solved by using the factorisation method and we also find the roots by using the formula sin⁡θ=−b±b2−4ac2a\sin \theta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}. While factorising we use sum product rule, the sum product rule is given as the product factors of the number c is equal to the sum of the factors which satisfies the value of b. The trigonometry is in the form of a quadratic equation. So we use the formula to simplify.