Question

Find \[x\]if \[\cos 3x + \sin \left( {2x - \dfrac{{7\pi }}{6}} \right) = - 2\]

Answer 1

Hint: Here, we have to solve for the variable by using the trigonometric equation. First, we will find the variable for two trigonometric functions. We will choose the variables which satisfy both the trigonometric functions. We will find the general term for the variable which satisfies for various integers. Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles.
Formula used:
We will use the following formulas:
1.Co-function Identities: \[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \]
2.Half Angle formula: \[\cos \theta + 1 = 2{\cos ^2}\dfrac{\theta }{2}\]
3.Double Angle formula: \[1 - \cos 2\theta = 2{\sin ^2}\theta \]
4.General term: \[\sin \theta = 0 \Leftrightarrow \theta = n\pi \] ; \[\cos \theta = 0 \Leftrightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{2}\]

Complete step-by-step answer:
We are given with a trigonometric equation.
\[\cos 3x + \sin \left( {2x - \dfrac{{7\pi }}{6}} \right) = - 2\]
Rewriting the equation, we have
\[\]
\[ \Rightarrow \cos 3x + 1 + \sin \left( {2x - \dfrac{{7\pi }}{6}} \right) + 1 = 0\]
Using the half angle formula, \[\cos \theta + 1 = 2{\cos ^2}\dfrac{\theta }{2}\], in the above equation, we get
\[ \Rightarrow 2\dfrac{{{{\cos }^2}3}}{2}x + 1 + \sin \left( {2x - \dfrac{2}{3}\pi - \dfrac{\pi }{2}} \right) = 0\]
\[ \Rightarrow 2{\cos ^2}\left( {\dfrac{3}{2}x} \right) + 1 - \sin \left( {\dfrac{\pi }{2} - \left( { - 2x + \dfrac{2}{3}\pi } \right)} \right) = 0\]
Now, using the Co-function identity,\[\sin \left( {\dfrac{\pi }{2} - \theta } \right) = \cos \theta \], we get
\[ \Rightarrow 2{\cos ^2}\left( {\dfrac{3}{2}x} \right) + 1 - \cos \left( {\dfrac{2}{3}\pi - 2x} \right) = 0\]
\[ \Rightarrow 2{\cos ^2}\left( {\dfrac{3}{2}x} \right) + 1 - \cos 2\left( {\dfrac{\pi }{3} - x} \right) = 0\]
Simplifying using the Double angle formula, \[1 - \cos 2\theta = 2{\sin ^2}\theta \], we get
\[ \Rightarrow 2{\cos ^2}\left( {\dfrac{3}{2}x} \right) + 2{\sin ^2}\left( {\dfrac{\pi }{3} - x} \right) = 0\]
Taking 2 common, we get
\[ \Rightarrow 2\left( {{{\cos }^2}\left( {\dfrac{3}{2}x} \right) + {{\sin }^2}\left( {\dfrac{\pi }{3} - x} \right)} \right) = 0\]
\[ \Rightarrow {\cos ^2}\left( {\dfrac{3}{2}x} \right) + {\sin ^2}\left( {\dfrac{\pi }{3} - x} \right) = 0\]
Equating the each term with 0 separately, we get
\[ \Rightarrow \cos \dfrac{{3x}}{2} = 0\] and \[\sin \left( {\dfrac{\pi }{3} - x} \right) = 0\]
\[ \Rightarrow \dfrac{3}{2}x = {\cos ^{ - 1}}(0)\] and \[\left( {\dfrac{\pi }{3} - x} \right) = {\sin ^{ - 1}}(0)\]
We know that the general term: \[\sin \theta = 0 \Leftrightarrow \theta = n\pi \] and \[\cos \theta = 0 \Leftrightarrow \theta = \left( {2n + 1} \right)\dfrac{\pi }{2}\] .
Using the general term of sine and cosine, we have
\[ \Rightarrow \dfrac{3}{2}x = \left( {2n + 1} \right)\dfrac{\pi }{2}\] and \[\left( {\dfrac{\pi }{3} - x} \right) = n\pi \]
Substituting the values of \[n\] , we get
\[ \Rightarrow \dfrac{3}{2}x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2}.....\] and \[\left( {\dfrac{\pi }{3} - x} \right) = 0,\pi ,2\pi ,.......\]
\[ \Rightarrow x = \dfrac{2}{3}\left( {\dfrac{\pi }{2},\dfrac{{3\pi }}{2},\dfrac{{5\pi }}{2},\dfrac{{7\pi }}{2}.....} \right)\] and \[x = \dfrac{\pi }{3} + \left( {0,\pi ,2\pi ,......} \right)\]
\[ \Rightarrow x = \dfrac{\pi }{3},\pi ,\dfrac{{5\pi }}{3},\dfrac{{7\pi }}{3}.....;x = \dfrac{\pi }{3},\dfrac{{4\pi }}{3},\dfrac{{7\pi }}{3},\dfrac{{10\pi }}{3}.......\]
Now, we will take out the common term for both the sine and cosine. Therefore, we get
\[ \Rightarrow x = \dfrac{\pi }{3},\dfrac{{7\pi }}{3}.....\]
Now, rewriting the equation using the general term, we have
\[ \Rightarrow x = \left( {6k + 1} \right)\dfrac{\pi }{3}\]
Therefore, \[x = \left( {6k + 1} \right)\dfrac{\pi }{3},k \in {\bf{Z}}\].

Note: We have to be careful in finding the general term of the variable. General solution is the solution consisting of all possible solutions of a trigonometric equation. All possible values which satisfy the given trigonometric equation are called solutions of the given trigonometric equation. Trigonometric equation has an infinite number of solutions.