Question

How can I check if the period of a trigonometric equation solution is \[n\pi \] or \[2n\pi \]?

Answer 1

Hint:In this question we have given a trigonometric equation for which we need to find how to find the period of that function in terms of\[\pi \].
A function \[f\left( x \right)\]is said to be periodic if there exists T > 0 such that \[f{\text{ }}\left( {x + T} \right){\text{ }} = {\text{ }}f\left( x \right)\]for all x in the domain of definition of\[f\left( x \right)\]. If T is the smallest positive real number such that is\[f{\text{ }}\left( {x + T} \right){\text{ }} = {\text{ }}f\left( x \right)\], then it is called the period of\[f\left( x \right)\].
The trigonometric functions such as \[sin,{\text{ }}cos{\text{ }}and{\text{ }}tan\] are periodic functions.

Complete step by step answer:We try to search for the solutions of the equation \[sin{\text{ }}\theta = 0\]other than\[\theta {\text{ }} = {\text{ }}0\].
By seeing the equation, one might straight away reach at the conclusion that \[\theta {\text{ }} = {\text{ }}0\] is the only solution.
But, in case of trigonometric equations, it is important to rule out all possibilities so as to reach at the correct solution.
This is possible only when OP coincides with OX or OX’.
When OP coincides with OX, \[\theta {\text{ }} = {\text{ }}0,{\text{ }} \pm {\text{ }}2\pi ,{\text{ }} \pm {\text{ }}4\pi {\text{ }}and{\text{ }} \pm {\text{ }}6\pi {\text{ }} \ldots \ldots \ldots {\text{ }}(1)\]
And when OP coincides with OX’\[\theta {\text{ }} = {\text{ }} \pm {\text{ }}\pi ,{\text{ }} \pm {\text{ }}3\pi ,{\text{ }} \pm {\text{ }}5\pi {\text{ }} \ldots \ldots \ldots {\text{ }}\left( 2 \right)\]
Thus from (1) and (2) it follows that at \[sin{\text{ }}\theta {\text{ }} = {\text{ }}0\]
\[ \Rightarrow \theta {\text{ }} = {\text{ }}n\pi ,{\text{ }}where{\text{ }}n{\text{ }} = {\text{ }}0,{\text{ }} \pm 1,{\text{ }} \pm 2 \ldots \ldots .\]
When OP coincides with OY,
\[ \Rightarrow \theta {\text{ }} = {\text{ }}\pi /2,{\text{ }}5\pi /2,{\text{ }}9\pi /2{\text{ }}or,{\text{ }} - 3\pi /2,{\text{ }} - 7\pi /2\;{\text{ }}.. \ldots \ldots \ldots {\text{ }}\left( 1 \right)\;\]
When OP coincides with OY’
\[ \Rightarrow \theta {\text{ }} = {\text{ }} - 3\pi /2,{\text{ }} - 7\pi /2{\text{ }}or,{\text{ }} - \pi /2,{\text{ }} - 5\pi /2\;\;\;\;\;\;{\text{ }} \ldots \ldots \ldots \ldots {\text{ }}\left( 2 \right)\;\]
Thus from (1) and (2) if follows that the general solution of\[cos{\text{ }}\theta {\text{ }} = {\text{ }}0{\text{ }}is{\text{ }}\theta {\text{ }}\left( {2n + 1} \right){\text{ }}\pi /2\], where \[n{\text{ }} = {\text{ }}0,{\text{ }} \pm 1,{\text{ }} \pm 2{\text{ }} \ldots \ldots \ldots \;\]
General solution of the equation\[\sin \theta = k\]
We know that when\[\sin \theta = k\], \[k\]has to be such that \[-1{\text{ }} \leqslant {\text{ }}k{\text{ }} \leqslant {\text{ }}1\;\]
We can always find some \[\alpha \in \left[ {-\pi /2,{\text{ }}\pi /2} \right]\]
As the function will be: \[sin{\text{ }}\left( { - \pi } \right)/2{\text{ }} = {\text{ }} - 1{\text{ }}\& {\text{ }}sin{\text{ }}\pi /2{\text{ }} = {\text{ }}1\], such that\[\sin \theta = k\], i.e. \[\alpha {\text{ }} = {\text{ }}si{n^{ - 1}}k\]
\[ \Rightarrow sin{\text{ }}\theta {\text{ }} = {\text{ }}sin{\text{ }}\alpha ,{\text{ }}\alpha \in \left[ {-\pi /2,{\text{ }}\pi /2} \right]\]
\[ \Rightarrow sin{\text{ }}\theta {\text{ }}-{\text{ }}sin{\text{ }}\alpha {\text{ }} = {\text{ }}0\]
\[ \Rightarrow 2{\text{ }}sin{\text{ }}\left\{ {\left( {\theta {\text{ }}-{\text{ }}\alpha } \right)/2} \right\}{\text{ }}cos{\text{ }}\left\{ {\theta {\text{ }} + {\text{ }}\alpha )/2} \right\}{\text{ }} = {\text{ }}0\]
From the above equation to be satisfied, either \[sin{\text{ }}\{ \left( {\theta {\text{ }}-{\text{ }}\alpha } \right)/2){\text{ }} = {\text{ }}0\;\]
On consequently \[\left( {\left( {\theta {\text{ }}-{\text{ }}\alpha } \right)/2} \right){\text{ }} = {\text{ }}integral{\text{ }}multiple{\text{ }}of{\text{ }}\pi \]
\[\therefore \theta {\text{ }}-{\text{ }}\alpha {\text{ }} = {\text{ }}2n\pi \]
Thus \[\theta {\text{ }} = {\text{ }}2n\pi {\text{ }} + {\text{ }}\alpha \]
From equation (1) and (2), we can conclude that\[\theta {\text{ }} = {\text{ }}n\pi {\text{ }} + {\text{ }}\left( {-1} \right)n{\text{ }}\alpha \], where n is integral multiple, is the general solution of the equation \[sin{\text{ }}\theta {\text{ }} = {\text{ }}k\;\]
If α is assumed to be the least positive value of θ which satisfies two given trigonometrically possible equations, then the general value of \[\theta \] will be\[2n\pi {\text{ }} + {\text{ }}\alpha \].

Additional information:
Following tips and steps will help you systematically solve trigonometric equations.
1. Try to reduce equation in terms of one single trigonometric ratio preferably\[sin{\text{ }}\theta {\text{ }}or{\text{ }}cos{\text{ }}\theta \].
If we have choice to convert a problem in sine or cosine and cosine form is convenient compared to sine form
2. Factorize the polynomial in terms of these ratios.
3. For LHS to be zero solve for each factor. And write down a general solution, for each factor based on the standard results that are derived earlier in this section.

Note:
We cannot define a unique method of solving trigonometric equations. In each case, success in solving a trigonometric equation depends, in particular, on the knowledge and application ability of trigonometric formulas and the practice of solving problems.
Many trigonometric formulas are true equalities for all the values of the variable’s appearing in them.
Never divide by any expression which is zero.