# Examples on Trigonometric Equations Set 2

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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

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