# Examples on Trigonometric Equations Set 3

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Example - 92

Solve for  \begin{align}x:\;|\cos x{|^{{{\sin }^2}x - \frac{3}{2}\sin x + \frac{1}{2}}}\;\; = 1\end{align}

Solution: Either $$|\cos x|\; = 1 \qquad \Rightarrow \qquad x = n\pi ,\;n \in \mathbb{Z}$$

or,

${\sin ^2}x - \frac{3}{2}\sin x + \frac{1}{2} = 0 \Rightarrow \quad \sin x = 1,\;\frac{1}{2}$

But  $$\sin x$$ cannot equal 1 since then becomes 0. Therefore,

\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\sin x = \frac{1}{2} \\ \Rightarrow \quad &x = 2n\pi + \frac{\pi }{6},\;n \in \mathbb{Z}\, \\ \end{align}

Example - 93

Solve for  $$x:\;\;\sec x + {\text{cosec}}\,x = c$$

Solution: Rearranging to  $$\sin x\;{\text{and}}\;\cos x,$$  we have:

$\sin x + \cos x = c\sin x\cos x$ squaring,

\begin{align} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&1 + \sin 2x = \frac{{{c^2}}}{4}{\sin ^2}2x\\ \Rightarrow \quad&\sin 2x = \frac{{2 \pm 2\sqrt {1 + {c^2}} }}{{{c^2}}} \\ \end{align}

for a real solution to exist,

\begin{align} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\left| {\frac{{2 \pm 2\sqrt {1 + {c^2}} }}{{{c^2}}}} \right| \in \left[ { - 1,\;1} \right] \\ \Rightarrow \quad &c \in \left( { - \infty ,\; - 2\sqrt 2 } \right]\; \cup \left[ {2\sqrt 2 ,\;\infty } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Verify!}}} \right) \\ \end{align}

Thus, given that c lies in the appropriate interval, the solution is

$x = \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{2 \pm 2\sqrt {1 + {c^2}} }}{{{c^2}}}} \right) + n\pi ,\;\;\;n \in \mathbb{Z}$

Example - 94

Solve for $$x$$ and $$y$$ :

${\tan ^2}(x + y) + {\cot ^2}(x + y) = - {x^2} - 2x + 1$

Solution: Note that

${({\text{LHS}})_{\min }} = {({\text{RHS}})_{\max }} = 2$

This must imply that

\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\tan (x + y) = \pm \;1, \quad - {x^2} - 2x + 1 = 2 \\ \Rightarrow \quad &x + y = n\pi + \frac{\pi }{4},\;n \in \mathbb{Z} \quad \Rightarrow \quad x = - 1 \\ \Rightarrow \quad &y = 1 + n\pi \pm \frac{\pi }{4},\;n \in \mathbb{Z} \\ \end{align}

Example - 95

Find the sum of all $$x$$ in the internal  $$[0,\;2\pi ]$$ such that

$3{\cot ^2}x + 8\cot x + 3 = 0$

Solution: Recall the graph of $$\cot x$$. It is a bijection on  $$(0,\;\pi )$$, and it is negative in \begin{align}\left( {\frac{\pi }{2},\;\pi } \right)\end{align}. This fact will be used in this problem.

Solving the quadratic $$3{y^2} + 8y + 3 = 0$$  yields the roots

${y_{1,\;2}} = \frac{{ - 8 \pm 2\sqrt 7 }}{6},\;{y_1}\;{y_2}\; = 1$

Since both $${y_1},\;{y_2}$$  are negative, there exists unique $${x_1},\;{x_2}$$  in \begin{align}\left( {\frac{\pi }{2},\;\pi } \right)\end{align} such that

${y_1} = \cot {x_1},\;{y_2} = \cot {x_2}$

Also,  $${x_1} + {x_2} \in (\pi ,\;2\pi )$$. We have

$\cot {x_1}\;\cot {x_2} = 1$

Given the constraints, this can happen if

${x_1} + {x_2} = \frac{{3\pi }}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Why ?}}} \right)$

Using a similar reasoning, there is another pair of reals $${x_3}$$ and  $${x_4}$$ in  $$(\pi ,\;2\pi )$$  which satisfies all the required constraints and such that

${x_3} + {x_4} = \frac{{7\pi }}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{Justify this too !}}} \right)$

These are all the possible roots of the given equation in the interval. Thus,

${x_1} + {x_2} + {x_3} + {x_4} = 5\pi$