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
