Examples on Trigonometric Equations Set 3

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,

\[{x_1} + {x_2} + {x_3} + {x_4} = 5\pi \]