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Question: If \(\sin \left( {3\theta } \right) = \cos \left( {2\theta } \right)\), then \(\theta \) is equal to...

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

Explanation

Solution

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 [0,2π][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⁡(3θ)=cos⁡(2θ)\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 [0,2π][0,2\pi ].
So, In order to solve the given trigonometric equationsin⁡(3θ)=cos⁡(2θ)\sin \left( {3\theta } \right) = \cos \left( {2\theta } \right) , we should first try to express all the expressions in terms of sin⁡θ\sin \theta .
Now, we know that sin⁡(3θ)=3sin⁡θ−4sin⁡3θ\sin \left( {3\theta } \right) = 3\sin \theta - 4{\sin ^3}\theta and cos⁡(2θ)=1−2sin⁡2θ\cos \left( {2\theta } \right) = 1 - 2{\sin ^2}\theta . So, substituting the value of these expressions in the trigonometric equation given to us.
⇒3sin⁡θ−4sin⁡3θ=1−2sin⁡2θ\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,
⇒4sin⁡3θ+1−2sin⁡2θ−3sin⁡θ=0\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,
⇒4sin⁡3θ−2sin⁡2θ−3sin⁡θ+1=0−−−−−−(1)\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⁡θ\sin \theta as 11 in equation (1)\left( 1 \right) , we get,
⇒4(1)3−2(1)2−3(1)+1=0\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,
⇒4−2−3+1=0\Rightarrow 4 - 2 - 3 + 1 = 0
⇒0=0\Rightarrow 0 = 0
Hence, we get that sin⁡θ=1\sin \theta = 1 is a root of the equation (1)\left( 1 \right). So, using the factor theorem, (sin⁡θ−1)\left( {\sin \theta - 1} \right) must be a factor.
Now, from equation (1)\left( 1 \right), we get,
⇒4sin⁡3θ−2sin⁡2θ−3sin⁡θ+1=0\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 (sin⁡θ−1)\left( {\sin \theta - 1} \right) common from all terms.
⇒4sin⁡3θ−4sin⁡2θ+2sin⁡2θ−2sin⁡θ−sin⁡θ+1=0\Rightarrow 4{\sin ^3}\theta - 4{\sin ^2}\theta + 2{\sin ^2}\theta - 2\sin \theta - \sin \theta + 1 = 0
Now, taking (sin⁡θ−1)\left( {\sin \theta - 1} \right) common from two consecutive terms, we get,
⇒4sin⁡2θ(sin⁡θ−1)+2sin⁡θ(sin⁡θ−1)−(sin⁡θ−1)=0\Rightarrow 4{\sin ^2}\theta (\sin \theta - 1) + 2\sin \theta (\sin \theta - 1) - (\sin \theta - 1) = 0
Now, taking (sin⁡θ−1)(\sin \theta - 1) common from all the terms of the equation, we get,
⇒(sin⁡θ−1)(4sin⁡2θ+2sin⁡θ−1)=0\Rightarrow \left( {\sin \theta - 1} \right)\left( {4{{\sin }^2}\theta + 2\sin \theta - 1} \right) = 0
Now, either (sin⁡θ−1)=0\left( {\sin \theta - 1} \right) = 0 or (4sin⁡2θ+2sin⁡θ−1)=0\left( {4{{\sin }^2}\theta + 2\sin \theta - 1} \right) = 0.
⇒sin⁡θ=1\Rightarrow \sin \theta = 1 or (4sin⁡2θ+2sin⁡θ−1)=0\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.
⇒(4sin⁡2θ+2sin⁡θ−1)=0\Rightarrow \left( {4{{\sin }^2}\theta + 2\sin \theta - 1} \right) = 0
Using the quadratic formula for solving the quadratic equation, we get,
⇒sin⁡θ=−2±22−4(4)(−1)2(4)\Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 4 \right)\left( { - 1} \right)} }}{{2\left( 4 \right)}}
⇒sin⁡θ=−2±4+168\Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{8}
⇒sin⁡θ=−2±208\Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {20} }}{8}
⇒sin⁡θ=−2±258\Rightarrow \sin \theta = \dfrac{{ - 2 \pm 2\sqrt 5 }}{8}
⇒sin⁡θ=−1±54\Rightarrow \sin \theta = \dfrac{{ - 1 \pm \sqrt 5 }}{4}
Hence, we get the values of sin⁡θ\sin \theta as −1+54\dfrac{{ - 1 + \sqrt 5 }}{4}, −1−54\dfrac{{ - 1 - \sqrt 5 }}{4} and 11.
Hence, the values of θ\theta that satisfy the given trigonometric equation sin⁡(3θ)=cos⁡(2θ)\sin \left( {3\theta } \right) = \cos \left( {2\theta } \right) are: sin⁡−1(−1+54){\sin ^{ - 1}}\left( {\dfrac{{ - 1 + \sqrt 5 }}{4}} \right), sin⁡−1(−1−54){\sin ^{ - 1}}\left( {\dfrac{{ - 1 - \sqrt 5 }}{4}} \right), and 11 .
So, the correct answer is “sin⁡−1(−1+54){\sin ^{ - 1}}\left( {\dfrac{{ - 1 + \sqrt 5 }}{4}} \right), sin⁡−1(−1−54){\sin ^{ - 1}}\left( {\dfrac{{ - 1 - \sqrt 5 }}{4}} \right), and 11”.

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.