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