Trigonometric Equations and Inequalities
3.0 Steps to solve trigonometric equations of the form$$a\cos \theta + b\sin \theta = c$$
3.0 Steps to solve trigonometric equations of the form$$a\cos \theta + b\sin \theta = c$$
In order to solve $$a\cos \theta + b\sin \theta = c$$ where $a,b,c \in R$ and $\left| c \right| \leqslant \sqrt {{a^2} + {b^2}} $.Following steps needed to be followed:
STEP $1$: Write the equation in the form $$a\cos \theta + b\sin \theta = c$$
STEP $2$: Put $$a = r\cos \alpha ...(1)$$ and $$b = r\sin \alpha ...(2)$$ Squaring $(1)$ and $(2)$ and then add them, we get $$r = \sqrt {{a^2} + {b^2}} $$ and $$\tan \alpha = \frac{b}{a}$$
STEP $3$: From STEP $2$, the equation now becomes, $$\begin{equation} \begin{aligned} r\cos (\theta - \alpha ) = c \\ \Rightarrow \cos (\theta - \alpha ) = \frac{c}{r} = \cos \beta \quad (say) \\\end{aligned} \end{equation} $$
STEP $4$: Solve the equations using general solutions mentioned above.
Example 1. Find the general solution of $$\sin 6\theta = 0$$
Solution: $$\begin{equation} \begin{aligned} \sin 6\theta = 0 \\ \Rightarrow 6\theta = n\pi ,\quad n \in Z \\ \Rightarrow \theta = \frac{{n\pi }}{6},\quad n \in Z \\\end{aligned} \end{equation} $$
Example 2. Find the general solution of $$\cos \frac{{6\theta }}{5} = 0$$
Solution: $$\begin{equation} \begin{aligned} \cos \frac{{6\theta }}{5} = 0 \\ \Rightarrow \frac{{6\theta }}{5} = \left( {2n + 1} \right)\pi ,\quad n \in Z \\ \Rightarrow \theta = \frac{{5\left( {2n + 1} \right)\pi }}{6},\quad n \in Z \\\end{aligned} \end{equation} $$
Example 3. Find the general solution of $${\tan ^2}\theta = 0$$
Solution: $$\begin{equation} \begin{aligned} {\tan ^2}\theta = 0 \\ \Rightarrow \tan \theta = 0, \\ \Rightarrow \theta = n\pi ,\quad n \in Z \\\end{aligned} \end{equation} $$
Example 4. Find the general solution of $$2\sin \theta + 1 = 0$$
Solution: $$\begin{equation} \begin{aligned} 2\sin \theta + 1 = 0 \\ \Rightarrow \sin \theta = \frac{{ - 1}}{2} \\ \Rightarrow \sin \theta = \sin \left( {\frac{{ - \pi }}{6}} \right)\quad [\sin \left( {\frac{{ - \pi }}{6}} \right) = \frac{{ - 1}}{2}] \\ \Rightarrow \theta = n\pi + {\left( { - 1} \right)^n}\left( {\frac{{ - \pi }}{6}} \right),\quad n \in Z \\ \Rightarrow \theta = n\pi + {\left( { - 1} \right)^{n + 1}}\left( {\frac{\pi }{6}} \right),\quad n \in Z \\\end{aligned} \end{equation} $$
Example 5. Find the general solution of $$\sqrt 3 \sec 2\theta = 2$$
Solution: $$\begin{equation} \begin{aligned} \sqrt 3 \sec 2\theta = 2 \\ \Rightarrow \sec 2\theta = \frac{2}{{\sqrt 3 }} \\ \Rightarrow \cos 2\theta = \frac{{\sqrt 3 }}{2}\quad [\cos x = \frac{1}{{\sec x}}] \\ \Rightarrow \cos 2\theta = \cos \frac{\pi }{6}\quad [\cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}] \\ \Rightarrow 2\theta = 2n\pi \pm \frac{\pi }{6},\quad n \in Z \\ \Rightarrow \theta = n\pi \pm \frac{\pi }{{12}},\quad n \in Z \\\end{aligned} \end{equation} $$
Example 6. Solve: $$7{\cos ^2}\theta + 3{\sin ^2}\theta = 4$$
Solution: $$\begin{equation} \begin{aligned} 7{\cos ^2}\theta + 3{\sin ^2}\theta = 4 \\ \Rightarrow 7(1 - {\sin ^2}\theta ) + 3{\sin ^2}\theta = 4\quad [Using\,{\cos ^2}\theta = 1 - {\sin ^2}\theta ] \\ \Rightarrow 7 - 7{\sin ^2}\theta + 3{\sin ^2}\theta = 4 \\ \Rightarrow - 4{\sin ^2}\theta = - 3 \\ \Rightarrow {\sin ^2}\theta = \frac{3}{4} \\ \Rightarrow {\sin ^2}\theta = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \\ \Rightarrow {\sin ^2}\theta = {\sin ^2}\frac{\pi }{3} \\ \Rightarrow \theta = n\pi \pm \frac{\pi }{3},\;n \in Z \\\end{aligned} \end{equation} $$
Example 7. Solve: $$2{\sin ^2}y + {\sin ^2}2y = 2$$
Solution: $$\begin{equation} \begin{aligned} 2{\sin ^2}y + {\sin ^2}2y = 2 \\ \Rightarrow {\left( {2\sin y\cos y} \right)^2} = 2 - 2{\sin ^2}y\quad [Using\sin 2y = 2\sin y\cos y] \\ \Rightarrow 4{\sin ^2}y{\cos ^2}y = 2{\cos ^2}y\quad [Using{\cos ^2}y = 1 - {\sin ^2}y] \\ \Rightarrow 4{\sin ^2}y{\cos ^2}y - 2{\cos ^2}y = 0 \\ \Rightarrow 2{\cos ^2}y\left( {2{{\sin }^2}y - 1} \right) = 0 \\ \Rightarrow 2{\cos ^2}y = 0\quad or,\;2{\sin ^2}y - 1 = 0 \\ \Rightarrow {\cos ^2}y = 0\quad or,\,{\sin ^2}y = \frac{1}{2} \\ \Rightarrow {\cos ^2}y = {\cos ^2}\frac{\pi }{2}\quad or,\;{\sin ^2}y = {\sin ^2}\frac{\pi }{4} \\ \Rightarrow y = n\pi \pm \frac{\pi }{2}\quad or,\;y = m\pi \pm \frac{\pi }{4},\quad where\;m,n \in Z \\\end{aligned} \end{equation} $$
Example 8. Solve: $$\sqrt 2 \sec \theta + \tan \theta = 1$$
Solution: $$\begin{equation} \begin{aligned} \sqrt 2 \sec \theta + \tan \theta = 1 \\ \Rightarrow \frac{{\sqrt 2 }}{{\cos }} + \frac{{\sin \theta }}{{\cos \theta }} = 1 \\ \Rightarrow \sqrt 2 + \sin \theta = \cos \theta \\ \Rightarrow \cos \theta - \sin \theta = \sqrt 2 \quad \ldots \left( 1 \right) \\ This\,equation\,is\,of\,the\,form\quad \,a\cos \theta + b\sin \theta = c \\ where,\,a = 1,b = - 1,c = \sqrt 2 \\ Let,a = 1 = r\cos \alpha \,and\,b = - 1 = r\sin \alpha \\ \therefore {a^2} + {b^2} = {1^2} + {\left( { - 1} \right)^2} = {r^2}{\cos ^2}\alpha + {r^2}{\sin ^2}\alpha \\ \Rightarrow {r^2} = 1 + 1 = 2 \\ \Rightarrow r = \sqrt 2 \\ And,\tan \alpha = \frac{{ - 1}}{1} = - 1 \\ \Rightarrow \alpha = {\tan ^{ - 1}}\left( { - 1} \right) = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4} \\ Rewriting\left( 1 \right), \\ \,r\cos \alpha \cos \theta - r\sin \alpha \sin \theta = \sqrt 2 \\ \Rightarrow r\left( {\cos \alpha \cos \theta - \sin \alpha \sin \theta } \right) = \sqrt 2 \\ \Rightarrow \sqrt 2 \cos \left( {\theta + \frac{{3\pi }}{4}} \right) = \sqrt 2 \\ \Rightarrow \cos \left( {\theta + \frac{{3\pi }}{4}} \right) = \cos 0 \\ \Rightarrow \left( {\theta + \frac{{3\pi }}{4}} \right) = 2n\pi \pm 0,n \in Z \\ \Rightarrow \theta = 2n\pi - \frac{{3\pi }}{4},n \in Z \\\end{aligned} \end{equation} $$
Example 9. Find all the solutions that exists between $0$ and $2\pi $ in the equation $${\sin ^4}\theta - 2{\sin ^2}\theta - 1 = 0$$
Solution: $$\begin{equation} \begin{aligned} {\sin ^4}\theta - 2{\sin ^2}\theta - 1 = 0 \\ Solving\;the\;quadratic\;equation, \\ {\sin ^2}\theta = \frac{{2 \pm \sqrt {4 + 4} }}{2} = 1 \pm \sqrt 2 \\ \Rightarrow {\sin ^2}\theta = 1 + \sqrt 2 \Rightarrow \sin \theta = \pm \sqrt {1 + \sqrt 2 }= \pm 1.55 \quad \left[ {out\,of\,range} \right] \\ or,{\sin ^2}\theta = 1 - \sqrt 2 < 0\quad [Not\,applicable] \\\end{aligned} \end{equation} $$
Hence, No solution exists.
Example 10. Find the number of solutions of the pair of equations $$2{\sin ^2}\theta - 2\cos 2\theta = 0$$ and $$2{\cos ^2}\theta - 3\sin \theta = 0$$ in the interval $\left[ {0,2\pi } \right]$.
Solution: $$\begin{equation} \begin{aligned} 2{\sin ^2}\theta - \cos 2\theta = 0 \\ \Rightarrow 1 - \cos 2\theta - \cos 2\theta = 0\quad [\cos 2\theta = 1 - 2{\sin ^2}\theta ] \\ \Rightarrow 1 - 2\cos 2\theta = 0 \\ \Rightarrow 1 = 2\cos 2\theta \\ \Rightarrow \cos 2\theta = \frac{1}{2} \\ \Rightarrow \cos 2\theta = \cos \frac{\pi }{3} \\ \Rightarrow 2\theta = \frac{\pi }{3},\frac{{ - \pi }}{3},\frac{{5\pi }}{3} \\ \Rightarrow \theta = \frac{\pi }{6},\frac{{ - \pi }}{6},\frac{{5\pi }}{6} \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} 2{\cos ^2}\theta - 3\sin \theta = 0 \\ \Rightarrow 2 - 2{\sin ^2}\theta - 3\sin \theta = 0 \\ \Rightarrow 2{\sin ^2}\theta + 3\sin \theta - 2 = 0 \\ \Rightarrow \sin \theta = \frac{{ - 3 \pm \sqrt {9 + 16} }}{4} \\ \Rightarrow \sin \theta = \frac{{ - 3 \pm 5}}{4} \\ \Rightarrow \sin \theta = - 2\;[Out\,of\,Range] \\ Or,\sin \theta = \frac{1}{2} \\ \Rightarrow \theta = \frac{\pi }{6},\frac{{5\pi }}{6} \\\end{aligned} \end{equation} $$
Since, both the equations have $$\frac{\pi }{6}$$ and $$\frac{{5\pi }}{6}$$ as common solutions for the pair of equations.
The number of solutions are TWO.
Example 11. The number of values of $x$ in the interval $\left[{0,2\pi}\right]$ satisfying the equation $3{\sin ^2}x - 7\sin x + 2 = 0$ is
a) 0
b) 2
c) 6
d) 10
Solution: $$\begin{equation} \begin{aligned} 3{\sin ^2}x - 7\sin x + 2 = 0 \\ \Rightarrow \sin x = \frac{{7 \pm \sqrt {49 - 24} }}{6} \\ \Rightarrow \sin x = \frac{{7 \pm 5}}{6} \\ \Rightarrow \sin x = \frac{{12}}{6} = 2\;[Out\;of\;Range] \\ Or,\sin x = \frac{2}{6} = \frac{1}{3} \\ \Rightarrow x = n\pi + {\left( { - 1} \right)^n}{\sin ^{ - 1}}\frac{1}{3},\;n = 0,1,2,3,4,5 \\ \Rightarrow x = {\sin ^{ - 1}}\frac{1}{3},\pi - {\sin ^{ - 1}}\frac{1}{3},2\pi + {\sin ^{ - 1}}\frac{1}{3},3\pi - {\sin ^{ - 1}}\frac{1}{3},4\pi + {\sin ^{ - 1}}\frac{1}{3},5\pi - {\sin ^{ - 1}}\frac{1}{3} \\\end{aligned} \end{equation} $$
So, in the above solution only two values lies in the required interval.
So, in the above solution only two values lies in the required interval.
There are $2$ possible values of $x$ in the interval $\left[{0,2\pi}\right]$
Correct option is (B).
Example 12. Solve: $$4{\cos ^2}x\sin x - 2{\sin ^2}x = 3\sin x$$
Solution: $$\begin{equation} \begin{aligned} 4{\cos ^2}x\sin x - 2{\sin ^2}x = 3\sin x \\ \Rightarrow 4\left( {1 - {{\sin }^2}x} \right)\sin x - 2{\sin ^2}x = 3\sin x \\ \Rightarrow 4\sin x - 4{\sin ^3}x - 2{\sin ^2}x = 3\sin x \\ \Rightarrow 4\sin x - 3\sin x - 4{\sin ^3}x - 2{\sin ^2}x = 0 \\ \Rightarrow \sin x - 4{\sin ^3}x - 2{\sin ^2}x = 0 \\ \Rightarrow \sin x\left( {1 - 4{{\sin }^2}x - 2\sin x} \right) = 0 \\ \Rightarrow \sin x = 0{\text{ }}or,{\text{ }}1 - 4{\sin ^2}x - 2\sin x = 0 \\ \Rightarrow x = n\pi {\text{ }}or,4{\sin ^2}x + 2\sin x - 1 = 0 \\ \Rightarrow x = n\pi {\text{ }}or,{\text{ }}\sin x = \frac{{ - 2 \pm \sqrt {4 + 16} }}{8} \\ \Rightarrow x = n\pi {\text{ }}or,{\text{ }}\sin x = \frac{{ - 1 + \sqrt 5 }}{4} = \sin \frac{\pi }{{10}}{\text{ }}or,{\text{ }}\sin x = \frac{{ - 1 - \sqrt 5 }}{4} = \sin \frac{{ - 3\pi }}{{10}} \\ \Rightarrow x = n\pi {\text{ }}or,{\text{ }}x = m\pi + {\left( { - 1} \right)^m}\frac{\pi }{{10}}{\text{ }}or,{\text{ }}x = k\pi + {\left( { - 1} \right)^k}\frac{{ - 3\pi }}{{10}} \\\end{aligned} \end{equation} $$
