Question

How do you find the exact value of the equation \[\cos \theta =3\sin 2\theta \] in the interval \[0\le \theta <2\pi \]?

Answer 1

Hint: In the given question, we have been asked to find the value of the given equation for the given interval i.e. \[0\le \theta <2\pi \]. In order to solve the question, first we need to simplify the equation by substituting by using the trigonometric identity i.e. \[\sin 2a=2\sin a\cos a\]. Then solving each factor by considering the interval and you will get the exact values of ‘x’.

Complete step by step solution:
We have given that,
\[\cos \theta =3\sin 2\theta \]
As using the trigonometric identity i.e. \[\sin 2a=2\sin a\cos a\], we get
\[\cos \theta =6\sin \theta \cos \theta \]
Rewrite the above equation as,
\[\cos \theta -6\sin \theta \cos \theta =0\]
Taking out the common factor from the above expression, we get
\[\cos \theta \left( 1-6\sin \theta \right)=0\]
Now solving,
\[\Rightarrow \cos \theta =0\]
On a trigonometric unit circle;
\[\Rightarrow \cos \theta =0\ when,\ 0\left\{ \theta =\dfrac{\pi }{2},\dfrac{3\pi }{2} \right.\]
Now solving,
\[\Rightarrow 1-6\sin \theta =0\]
Simplifying the above,
\[\Rightarrow \sin \theta =\dfrac{1}{6}\ \]
\[\Rightarrow \theta ={{\sin }^{-1}}\dfrac{1}{6}\]
Using the calculator for the value of given arcsin,
\[\Rightarrow \theta =0.1674\]
Since the function sine is positive in first and second quadrants.
Thus, for the second solution subtract the reference angle from \[\pi \].
Therefore,
\[\Rightarrow \theta =3.1415-0.1674=2.974\]
\[\Rightarrow \theta =2.974\]
\[\Rightarrow \sin \theta =\dfrac{1}{6}\ when,\ \left\{ \theta =0.167,2.974 \right.\]

Therefore, The values of \[x\] are\[\dfrac{\pi }{2},\ \dfrac{3\pi }{2},\ 0.167,\ 2.974\].

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric ratios as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.