Question

The solution of the trigonometric equation \[\cos 3\theta =4\cos \theta \cos \left( \theta +x \right)\cos \left( \theta -x \right)\] such that \[0<\theta <\pi \] is
A. \[k\pi +\dfrac{\pi }{3}\]
B. \[2k\pi \]
C. \[k\pi -\dfrac{\pi }{3}\]
D. None of these

Answer 1

Hint: Use the trigonometric identity \[\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta \] and \[\cos \left( a\pm b \right)=\cos a\cos b\mp \sin a\sin b\] to simplify the given trigonometric equation and find the value of x and write the general equation of trigonometric function.

Complete step-by-step answer:
We have the trigonometric equation \[\cos 3\theta =4\cos \theta \cos \left( \theta +x \right)\cos \left( \theta -x \right).....\left( 1 \right)\]. We have to find the value of x which satisfies the given trigonometric equation.
We know the trigonometric identity \[\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta .....\left( 2 \right)\] and \[\cos \left( a\pm b \right)=\cos a\cos b\mp \sin a\sin b\].
Thus, we have \[\cos \left( \theta +x \right)=\cos \theta \cos x-\sin \theta \sin x.....\left( 3 \right)\] and \[\cos \left( \theta -x \right)=\cos \theta \cos x+\sin \theta \sin x.....\left( 4 \right)\].
Substituting all the equations in equation (1) , we have \[4{{\cos }^{3}}\theta -3\cos \theta =4\cos \theta \left( \cos \theta \cos x-\sin \theta \sin x \right)\left( \cos \theta \cos x+\sin \theta \sin x \right)....\left( 5 \right)\].
We know the algebraic identity \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\].
Substituting \[a=\cos \theta \cos x,b=\sin \theta \sin x\] in the above equation, we have \[\left( \cos \theta \cos x-\sin \theta \sin x \right)\left( \cos \theta \cos x+\sin \theta \sin x \right)={{\cos }^{2}}\theta {{\cos }^{2}}x-{{\sin }^{2}}\theta {{\sin }^{2}}x.....\left( 6 \right)\].
Substituting equation (6) in equation (5), we have \[4{{\cos }^{3}}\theta -3\cos \theta =4\cos \theta \left( {{\cos }^{2}}\theta {{\cos }^{2}}x-{{\sin }^{2}}\theta {{\sin }^{2}}x \right).....\left( 7 \right)\].
We know that \[{{\cos }^{2}}a+{{\sin }^{2}}a=1\].
Thus, we have \[{{\sin }^{2}}a=1-{{\cos }^{2}}a\].
Rewriting equation (7), we have \[4{{\cos }^{3}}\theta -3\cos \theta =4\cos \theta \left( {{\cos }^{2}}\theta {{\cos }^{2}}x-\left( 1-{{\cos }^{2}}\theta \right)\left( 1-{{\cos }^{2}}x \right) \right)\].
Simplifying the above equation, we have \[4{{\cos }^{3}}\theta -3\cos \theta =4\cos \theta \left( {{\cos }^{2}}\theta {{\cos }^{2}}x-1+{{\cos }^{2}}x+{{\cos }^{2}}\theta -{{\cos }^{2}}\theta {{\cos }^{2}}x \right)\].
Thus, we have \[4{{\cos }^{3}}\theta -3\cos \theta =4\left( {{\cos }^{3}}\theta {{\cos }^{2}}x-\cos \theta +\cos \theta {{\cos }^{2}}x+{{\cos }^{3}}\theta -{{\cos }^{3}}\theta {{\cos }^{2}}x \right)\].
So, we have \[4{{\cos }^{3}}\theta -3\cos \theta =4{{\cos }^{3}}\theta +4\cos \theta ({{\cos }^{2}}x-1)\].
Thus, we have \[\cos \theta (4{{\cos }^{2}}x-1)=0\].
So, we have \[{{\cos }^{2}}x=\dfrac{1}{4}\].
Taking the square root on both sides, we have \[\cos x=\pm \dfrac{1}{2}\].
Thus, for \[\cos x=\dfrac{1}{2}\], we have \[x=2n\pi \pm \dfrac{\pi }{3}\].
For \[\cos x=\dfrac{-1}{2}\], we have \[x=2n\pi \pm \dfrac{2\pi }{3}=\left( 2n+1 \right)\pi \pm \dfrac{\pi }{3}\].
Hence, the value of x which satisfies the given equation \[\cos 3\theta =4\cos \theta \cos \left( \theta +x \right)\cos \left( \theta -x \right)\] is \[x=n\pi \pm \dfrac{\pi }{3}\], which is option A, C.
Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

Note: One must be careful while finding the values of x which satisfy the given equation. We need to consider both positive and negative roots of the equation \[{{\cos }^{2}}x=\dfrac{1}{4}\]; otherwise, we won’t get a correct answer.