Question

If $\sin \left( {3\theta } \right) = \cos \left( {2\theta } \right)$, then $\theta $ is equal to ?

Answer 1

Hint: The given question involves solving a trigonometric equation and finding the value of angle x that satisfies the given equation and lies in the range of $[0,2\pi ]$. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities.

Complete step-by-step answer:
In the given problem, we have to solve the trigonometric equation $\sin \left( {3\theta } \right) = \cos \left( {2\theta } \right)$ and find the values of x that satisfy the given equation and lie in the range of $[0,2\pi ]$.
So, In order to solve the given trigonometric equation$\sin \left( {3\theta } \right) = \cos \left( {2\theta } \right)$ , we should first try to express all the expressions in terms of $\sin \theta $.
Now, we know that $\sin \left( {3\theta } \right) = 3\sin \theta - 4{\sin ^3}\theta $ and $\cos \left( {2\theta } \right) = 1 - 2{\sin ^2}\theta $. So, substituting the value of these expressions in the trigonometric equation given to us.
$ \Rightarrow 3\sin \theta - 4{\sin ^3}\theta = 1 - 2{\sin ^2}\theta $
Shifting all the terms to the right side of the equation, we get,
\[ \Rightarrow 4{\sin ^3}\theta + 1 - 2{\sin ^2}\theta - 3\sin \theta = 0\]
Arranging all the terms and simplifying the expression further, we get,
\[ \Rightarrow 4{\sin ^3}\theta - 2{\sin ^2}\theta - 3\sin \theta + 1 = 0 - - - - - - (1)\]
Now, we can see that on substituting the value of \[\sin \theta \] as $1$ in equation $\left( 1 \right)$ , we get,
\[ \Rightarrow 4{\left( 1 \right)^3} - 2{\left( 1 \right)^2} - 3\left( 1 \right) + 1 = 0\]
Now, we know that the value of any power of one is one. Hence, we get,
\[ \Rightarrow 4 - 2 - 3 + 1 = 0\]
\[ \Rightarrow 0 = 0\]
Hence, we get that \[\sin \theta = 1\] is a root of the equation $\left( 1 \right)$. So, using the factor theorem, $\left( {\sin \theta - 1} \right)$ must be a factor.
Now, from equation $\left( 1 \right)$, we get,
\[ \Rightarrow 4{\sin ^3}\theta - 2{\sin ^2}\theta - 3\sin \theta + 1 = 0\]
Now, manipulating the terms in the equation so as to get $\left( {\sin \theta - 1} \right)$ common from all terms.
\[ \Rightarrow 4{\sin ^3}\theta - 4{\sin ^2}\theta + 2{\sin ^2}\theta - 2\sin \theta - \sin \theta + 1 = 0\]
Now, taking $\left( {\sin \theta - 1} \right)$ common from two consecutive terms, we get,
\[ \Rightarrow 4{\sin ^2}\theta (\sin \theta - 1) + 2\sin \theta (\sin \theta - 1) - (\sin \theta - 1) = 0\]
Now, taking \[(\sin \theta - 1)\] common from all the terms of the equation, we get,
\[ \Rightarrow \left( {\sin \theta - 1} \right)\left( {4{{\sin }^2}\theta + 2\sin \theta - 1} \right) = 0\]
Now, either \[\left( {\sin \theta - 1} \right) = 0\] or \[\left( {4{{\sin }^2}\theta + 2\sin \theta - 1} \right) = 0\].
\[ \Rightarrow \sin \theta = 1\] or \[\left( {4{{\sin }^2}\theta + 2\sin \theta - 1} \right) = 0\]
Now, we have to factorize the quadratic equation in sine function in order to solve the equation.
\[ \Rightarrow \left( {4{{\sin }^2}\theta + 2\sin \theta - 1} \right) = 0\]
Using the quadratic formula for solving the quadratic equation, we get,
\[ \Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 4 \right)\left( { - 1} \right)} }}{{2\left( 4 \right)}}\]
\[ \Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{8}\]
\[ \Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {20} }}{8}\]
\[ \Rightarrow \sin \theta = \dfrac{{ - 2 \pm 2\sqrt 5 }}{8}\]
\[ \Rightarrow \sin \theta = \dfrac{{ - 1 \pm \sqrt 5 }}{4}\]
Hence, we get the values of \[\sin \theta \] as \[\dfrac{{ - 1 + \sqrt 5 }}{4}\], \[\dfrac{{ - 1 - \sqrt 5 }}{4}\] and $1$.
Hence, the values of $\theta $ that satisfy the given trigonometric equation $\sin \left( {3\theta } \right) = \cos \left( {2\theta } \right)$ are: \[{\sin ^{ - 1}}\left( {\dfrac{{ - 1 + \sqrt 5 }}{4}} \right)\], \[{\sin ^{ - 1}}\left( {\dfrac{{ - 1 - \sqrt 5 }}{4}} \right)\], and $1$ .
So, the correct answer is “\[{\sin ^{ - 1}}\left( {\dfrac{{ - 1 + \sqrt 5 }}{4}} \right)\], \[{\sin ^{ - 1}}\left( {\dfrac{{ - 1 - \sqrt 5 }}{4}} \right)\], and $1$”.

Note: Such questions that involve solving such trigonometric equations should be solved with care as it involves many topics of mathematics. One must have a strong grip over the concepts of trigonometry and algebra. Care should be taken while handling the calculations so as to be sure of the answer of the question.