Examples on Trigonometric Equations Set 1

In this section, nothing new is going to be covered. We’ll just be going through a few examples on how trig. equations are solved.

The basic purpose of solving any trig. equation is finding its general solution; since the trigonometric functions are periodic, there can be multiple (or we infinite) solutions to a given trig. equation.

Example - 86

Solve for  \(\theta :\sqrt 2 \sec \theta  + \tan \theta  = 1\)

Solution:

\[\begin{align}&\frac{{\sqrt 2 }}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }} = 1  \\   \Rightarrow\quad &\sqrt 2  + \sin \theta  = \cos \theta \\   \Rightarrow\quad &\frac{1}{{\sqrt 2 }}\cos \theta  - \frac{1}{{\sqrt 2 }}\sin \theta  = 1  \;\;\;\;\; \Rightarrow \quad  \;\;\;\;\cos \left( {\frac{\pi }{4} + \theta } \right) = 1 \\   \Rightarrow\quad&\frac{\pi }{4} + \theta  = 2n\pi ,\;n \in \mathbb{Z}\;\;\;\qquad({\text{How}}?) \\   \Rightarrow\quad &\theta  = 2n\pi  - \frac{\pi }{4},\;n \in \mathbb{Z} \\ \end{align} \]

We see that  \(\theta \)  can have an infinite number of values.

Example - 87

Solve for  \(\begin{align}x:{\sin ^2}x + \frac{{{{\sin }^2}3x}}{4} = \sin x{\sin ^2}3x\end{align}\)

Solution:

\[\begin{align}&{\sin ^2}x - \sin x{\sin ^2}3x + \frac{{{{\sin }^2}3x}}{4} = 0 \\   \Rightarrow  \quad &{\left( {\sin x - \frac{1}{2}{{\sin }^2}3x} \right)^2} + \frac{{{{\sin }^2}3x}}{4} - \frac{{{{\sin }^4}3x}}{4} = 0 \quad ({\text{How}}?) \\   \Rightarrow  \quad& {\left( {\sin x - \frac{1}{2}{{\sin }^2}3x} \right)^2} + {\left( {\frac{{\sin 6x}}{4}} \right)^2} = 0  \\   \Rightarrow  \quad &\sin x = \frac{1}{2}{\sin ^2}3x\;{\text{AND}}\;\sin 6x = 0  \\   &\qquad \qquad \Rightarrow  \quad 6x = n\pi ,\;n \in \mathbb{Z}  \\   &\qquad \qquad \Rightarrow \quad x = n\pi /6,\;n \in \mathbb{Z}  \\ \end{align} \]

If  \(\begin{align}x = \frac{{n\pi }}{6},\;{\text{then}}\;3x = \frac{{n\pi }}{2}\end{align}\)

\[\begin{align}&\sin 3x = \left\{ \begin{gathered}  0,\;{\text{if}}\;n\;{\text{is}}\;{\text{even}}  \\  {\text{1,}}\;{\text{if}}\;n\;{\text{is}}\;{\text{odd}} \\ \end{gathered}  \right\} \\   \Rightarrow  \quad &\sin x = 0\;{\text{or}}\;\frac{1}{2} \\   \Rightarrow  \quad &\pi  = n\pi \;{\text{or}}\;n\pi  + {( - 1)^n}\frac{\pi }{6},\;n \in \mathbb{Z} \\ \end{align} \]

 Example - 88

Solve for  \(x  :|\cos x - 2\sin 2x - \cos 3x| = 1 - 2\sin x - \cos 2x\)

Solution: Combining  \(\cos x - \cos 3x\) on the LHS, and writing  \(1 - \cos 2x = 2{\sin ^2}x\) on the RHS, we’ll have:

\[|2\sin 2x(\sin x - 1)|\; = 2\sin x(\sin x - 1)\]

Since \(\sin x - 1 \leqslant 0,\)  we have

\[(\sin x - 1)\left\{ {\sin x + |\sin 2x|} \right\} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{How?}}} \right)\]

This means that