Question

Solve the given quadratic equation \[2{\cos ^2}x + 3\sin x = 0\]

Answer 1

Hint: Here, we have to solve the given equation and find the value of \[x\]. First, we will use the trigonometric identities to convert the given equation into the form as a quadratic equation. Then, we will solve the equation by factoring method. Again, we will use the trigonometric identities converting the variable to trigonometric function and using periodicity identities, we will obtain the value of the variable. Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles.

Formula used: We will use the following formulas:
Trigonometric Identity: \[{\sin ^2}x + {\cos ^2}x = 1 \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x\]
Periodicity Identities: \[\sin (\pi + A) = - \sin A;\sin (2\pi - A) = - \sin A;\]

Complete step-by-step answer:
We are given with a trigonometric function to solve for the variable.
\[2{\cos ^2}x + 3\sin x = 0\]
We know that \[{\sin ^2}x + {\cos ^2}x = 1\], so rewriting this identity, we get
\[{\cos ^2}x = 1 - {\sin ^2}x\]
Substituting \[{\cos ^2}x = 1 - {\sin ^2}x\] in the given equation, we get
\[ \Rightarrow 2(1 - {\sin ^2}x) + 3\sin x = 0\]
Multiplying the terms, we get
\[ \Rightarrow 2 - 2{\sin ^2}x + 3\sin x = 0\]
Rewriting the above equation, we get
\[ \Rightarrow 2{\sin ^2}x - 3\sin x - 2 = 0\] …………………………….\[\left( 1 \right)\]
Now, Let \[\sin x = y\]. Therefore,
\[ \Rightarrow 2{y^2} - 3y - 2 = 0\]
By using factoring method for Quadratic equation, we have
\[ \Rightarrow 2{y^2} - 4y + y - 2 = 0\]
Now, grouping into terms, we get
\[ \Rightarrow 2y\left( {y - 2} \right) + 1\left( {y - 2} \right) = 0\]
Now, again by grouping into terms, we have
\[ \Rightarrow \left( {2y - 1} \right)\left( {y - 2} \right) = 0\]
Now, by using the Zero-Product Property, we have
\[ \Rightarrow \left( {2y + 1} \right) = 0\] or \[\left( {y - 2} \right) = 0\]
Rewriting the equation, we have
\[ \Rightarrow 2y = - 1\] or \[y = 2\]
\[ \Rightarrow y = \dfrac{{ - 1}}{2}\] or \[y = 2\]
Now substituting \[y = \sin x\] in the above equations, we get
\[ \Rightarrow \sin x = \dfrac{{ - 1}}{2}\] or \[\sin x = 2\]
Since sine lies between \[ - 1\]to \[ + 1\] and \[\sin x = 2\] is greater than \[ + 1\], so we will not take \[\sin x = 2\].
Now,
\[\sin x = \dfrac{{ - 1}}{2}\]
\[ \Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)\]
\[\sin x\]is negative in the third and the fourth quadrant.
Using the periodicity identities \[\sin (\pi + A) = - \sin A\] and \[\sin (2\pi - A) = - \sin A\], we have
\[ \Rightarrow x = \pi + \dfrac{\pi }{6}\] or \[x = 2\pi - \dfrac{\pi }{6}\]
By cross multiplying, we have
\[ \Rightarrow x = \dfrac{{6\pi }}{6} + \dfrac{\pi }{6}\] or \[x = \dfrac{{12\pi }}{6} - \dfrac{\pi }{6}\]
Adding the like fractions, we have
\[ \Rightarrow x = \dfrac{{6\pi + \pi }}{6}\] or \[x = \dfrac{{12\pi - \pi }}{6}\]
\[ \Rightarrow x = \dfrac{{7\pi }}{6}\] or \[x = \dfrac{{11\pi }}{6}\]
Therefore, the values are \[x = \dfrac{{7\pi }}{6}\] and \[\dfrac{{11\pi }}{6}\].

Note: We have used the property of Zero-Product Property in Factoring method. Zero-Product Property states that If we multiply two (or more) things together and the result is equal to zero.Then we know that at least one of those things that we multiplied must also have been equal to zero. The only way for us to get zero when we multiply two (or more) factors together is for one of the factors to have been zero. We should be clear of Periodicity identities too.