Examples on Trigonometric Equations Set 2

Go back to  'Trigonometry'

 Example - 89

Solve for  \(\alpha \)  and  \(\beta \) :

\[12\sin \alpha  + 5\cos \alpha  = 2{\beta ^2} - 8\beta  + 21\]

Solution: The LHS can be written as

\[\begin{align}&{\text{LHS}} = 13\left( {\frac{{12}}{{13}}\sin \alpha  + \frac{5}{{13}}\cos \alpha } \right)\\  \,\,\,\,\,\,\,\,\,\,\, &\qquad= 13\sin (\alpha  + \phi )\,\,{\text{Where}}\,\,{\text{ }}\tan \phi  = \frac{5}{{12}} \\   \Rightarrow\quad  &{({\text{LHS}})_{\max }} = 13 \\\end{align}\]

The RHS can be written as

\[\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&{\text{RHS}} = 2{(\beta  - 2)^2} + 13 \geqslant 13 \\   \Rightarrow \quad  &{({\text{RHS}})_{\min }} = 13  \\ \end{align} \]

The only way the two can be equal are if both are equal to 13:

\[\begin{align} & \sin (\alpha  + \phi ) = 1 \quad{\text{AND}}\quad{\text{RHS}} = 13  \\ \Rightarrow  \quad &\alpha  + \phi  = 2n\pi  + \frac{\pi }{2}   \quad\Rightarrow\quad \beta  = 2  \\   \Rightarrow  \quad &\alpha  = 2n\pi  + \frac{\pi }{2} - \phi \\\end{align} \]

where \(\begin{align}\phi  = {\tan ^{ - 1}}\frac{5}{{12}}\end{align}\)

Example - 90

For what values of  \(\lambda \)  does the equation

\[\frac{{\lambda \cos 2x}}{{2\cos 2x - 1}} = \frac{{\lambda  + \sin x}}{{({{\cos }^2} - 3{{\sin }^2}x)\tan x}}\]

have real solutions?

Solution: We note that  \({\cos ^2}x - 3{\sin ^2}x = 4{\cos ^2}x - 3 = 2\cos 2x - 1\)

The equation is defined if

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\cos 2x - 1\; \ne 0  \;\;\;\;\;\;\;{\text{AND}}\quad  \tan x \ne 0  \\   \Rightarrow\quad & x \ne n\pi  \pm \frac{\pi }{6},\;n \in \mathbb{Z}   \quad\Rightarrow  \quad x \ne n\pi ,\;n \in \mathbb{Z}  \\ \end{align} \]

Also, should itself be defined   \[ \Rightarrow  \quad x \ne (2n + 1)\frac{\pi }{2},\;n \in \mathbb{Z}\]

Given that these constraints are satisfied, the equation reduces to

\[\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\lambda \sin x = \lambda  + \sin x\,  \\ \Rightarrow  \quad &\sin x = \frac{\lambda }{{\lambda  - 1}} \\ \end{align} \]

\[\begin{align}   \Rightarrow\quad  &- 1 \leqslant \frac{\lambda }{{\lambda  - 1}} \leqslant \;1 \\   \Rightarrow\quad  &\frac{\lambda }{{\lambda  - 1}} \leqslant \;1  \quad {\text{and}} \quad  \frac{\lambda }{{\lambda  - 1}} \geqslant \; - 1 \\   \Rightarrow  \quad &\frac{1}{{\lambda  - 1}} \leqslant \;0 \quad {\text{and}} \quad \frac{{2\lambda  - 1}}{{\lambda  - 1}} \geqslant \;0 \\   \Rightarrow  \quad &\lambda  < 1 \quad{\text{and}}\quad  \lambda  \leqslant \frac{1}{2}\;\;{\text{or}}\;\lambda  \geqslant \;1  \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\quad\Rightarrow \quad \lambda \; \leqslant \;\frac{1}{2} \\ \end{align} \]

If \(\begin{align}\lambda \; = \frac{1}{2}\end{align}\), the equation becomes \(\begin{align}\sin x =  - 1\end{align}\)

\[ \Rightarrow  \quad x = ....,\;\frac{{ - \pi }}{2},\;\frac{{3\pi }}{2},\;\frac{{7\pi }}{2},....\]

But this is not allowed since at these values, becomes undefined. Thus, \(\begin{align}\lambda  < \frac{1}{2}\end{align}\)

Also, since  \(\begin{align}x \ne n\pi ,\;nx \pm \frac{\pi }{6}\end{align}\)

\[\begin{align}  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\sin x = \frac{\lambda }{{\lambda  - 1}} \ne 0,\;\frac{1}{2},\; - \frac{1}{2} \Rightarrow \lambda  \ne 0,\; - 1,\;\frac{1}{3} \\   \Rightarrow\quad &\lambda  \in \left( { - \infty ,\;\frac{1}{2}} \right)\backslash \left\{ { - 1,\;0,\;\frac{1}{3}} \right\}\\ \end{align} \]

Example - 91

Solve for  \(\theta :\;\;\cos 4\theta  + \sin 5\theta  = 2\)

Solution: It should be immediately evident that since the maximum value of both the sin and cos functions is 1, the given equality can only hold if

\[\begin{align}  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\cos 4\theta  = 1 \quad {\text{AND}} \quad \sin 5\theta  = 1  \\   \Rightarrow  \quad &4\theta  = 2n\pi  \quad {\text{AND}} \quad 5\theta  = 2n\pi  + \frac{\pi }{2} \\   \Rightarrow  \quad &\theta  = \frac{{n\pi }}{2} \quad {\text{AND}} \quad \theta  = \frac{{2m\pi }}{5} + \frac{\pi }{{10}}\;\;{\text{for}}\;{m_1}n \in \mathbb{Z} \\ \end{align} \]

Therefore, we have to find \(m, n\) such that

\[\begin{align}&\frac{{n\pi }}{2} = \frac{{2m\pi }}{5} + \frac{\pi }{{10}}  \\   \Rightarrow  \quad &5n = 4m + 1 \\   \Rightarrow  \quad &5(n - 1) = 4(m - 1)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{say}}} \right)\\   \Rightarrow  \quad &\frac{{n - 1}}{4} = \frac{{m - 1}}{5} = \lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{say}}} \right)\,\,{\text{Where }}\,\,\lambda  \in \mathbb{Z}  \\   \Rightarrow  \quad &n = 4\lambda  + 1,\;m = 5\lambda  + 1,\;\lambda  \in \mathbb{Z} \\ \end{align} \]

The solutions to the given equation can thus be written as

\[\theta  = (4\lambda  + 1)\frac{\pi }{2},\;\;\lambda  \in \mathbb{Z}\]

Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum