Question

If \[2{\sin ^2}\theta = 3\cos \theta \], where\[0 \le \theta \le 2\pi \], then\[\theta = \]
А. \[\frac{\pi }{6},\frac{{7\pi }}{6}\]
В. \[\frac{\pi }{3},\frac{{5\pi }}{3}\]
C. \[\frac{\pi }{3},\frac{{7\pi }}{3}\]
D. None of these

Answer 1

Hints
We will employ the trigonometric identity in the equation above, which is provided as follows:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\] We will solve the quadratic equation in cos after using the identity to obtain the answer to the equation.
Formula used:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Complete step-by-step solution
We have been given the equation,
\[2{\sin ^2}\theta = 3\cos \theta \]
As it is known that,
\[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]
By using the above equation, we obtain
\[2\left( {1 - {{\cos }^2}\theta } \right) = 3\cos \theta \]
\[2 - 2{\cos ^2}\theta = 3\cos \theta \]
The following results are obtained by transferring all the terms from the left to the right side of the equation:
\[0 = 3\cos \theta + 2{\cos ^2}\theta - 2\]
Simplify further to obtain the less complicated form of the equation:
\[2{\cos ^2}\theta + 3\cos \theta - 2\]
In order to factorize the above equation in \[\cos \theta \], we can write \[3cos\theta \] in the form of\[\left( {4\cos \theta - \cos \theta } \right)\]:
\[2{\cos ^2}\theta + 4\cos \theta - \cos \theta - 2 = 0\]
The above equation can also be written as,
\[2\cos \theta (\cos \theta + 2) - 1(\cos \theta + 2) = 0\]
Now after taking \[\cos \theta \] as common, we obtain the resultant equation as
\[(\cos \theta + 2)(2\cos \theta - 1) = 0\]
So, the expression looks like
\[ \Rightarrow \cos \theta + 2 = 0\]
And the other equation looks like
\[2\cos \theta - 1 = 0\]
As it is known that the range of\[\cos \theta \] is \[[ - 1,1]\].
\[ \Rightarrow \cos \theta \ne - 2\]
So we have only the equation left is,
\[2\cos \theta - 1 = 0\].
After that, we have to solve for\[\cos \theta \]:
\[ \Rightarrow \cos \theta = \frac{1}{2}\]
As we know that the general solution of\[\cos \theta = \cos \alpha \] is given by \[\theta = 2n\pi \pm \alpha \] where 'n' is any integer.
\[\cos \theta = \cos \frac{\pi }{3}\]
Now, solve for theta to make the equation less complicated:
\[\theta = 2n\pi \pm \frac{\pi }{3}\]
It is been given to us that \[0 \le \theta \le 2\pi \].
\[ \Rightarrow \theta = \frac{\pi }{3},\frac{{5\pi }}{3}\]
Therefore, the solution for the given equations \[2{\sin ^2}\theta = 3\cos \theta \] are \[\theta = \frac{\pi }{3}\] and\[\theta = \frac{{5\pi }}{3}\].
Hence, the option B is correct.
Note
Given that\[0 \le \theta \le 2\pi \], student should take care while determining the equation's precise solution. Also, exercise caution when resolving the quadratic equation in cos, as there is a possibility that you could err when eliminating the common terms. By creating the graph of \[\cos \theta = \frac{1}{2}\] for\[0 \le \theta \le 2\pi \], we can also determine the precise solution.