Question

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

Answer 1

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

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 \[\sin \theta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. So the equation is written as \[2{\sin ^2}\theta + 3\sin \theta + 1 = 0\].
In general, the quadratic equation is represented as \[a{x^2} + bx + c = 0\], 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
\[\sin \theta = \dfrac{{ - 3 \pm \sqrt {{3^2} - 4(2)(1)} }}{{2(2)}}\]
On simplifying the terms, we have
\[ \Rightarrow \sin \theta = \dfrac{{ - 3 \pm \sqrt {9 - 8} }}{4}\]
Now subtract 8 from 9 we get
\[ \Rightarrow \sin \theta = \dfrac{{ - 3 \pm \sqrt 1 }}{4}\]
The number 1 is a perfect square so we can take out from square root we have
\[ \Rightarrow \sin \theta = \dfrac{{ - 3 \pm 1}}{4}\]
Therefore, we have \[\sin {\theta _1} = \dfrac{{ - 3 + 1}}{4} = - \dfrac{2}{4} = - \dfrac{1}{2}\] or \[\sin {\theta _2} = \dfrac{{ - 3 - 1}}{4} = \dfrac{{ - 4}}{4} = - 1\]
The value of \[\theta \] can be determined by
 \[{\theta _1} = {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\] and \[{\theta _2} = {\sin ^{ - 1}}( - 1)\]
So we have a table for the trigonometry ratio sine for the standard angles.

Angle	0	30	45	60	90
sine	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	1

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

Note: The quadratic equation can be solved by using the factorisation method and we also find the roots by using the formula \[\sin \theta = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. 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.