# Examples on Trigonometric Equations Set 1

Go back to  'Trigonometry'

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