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

How can I check if the period of a trigonometric equation solution is nπn\pi or 2nπ2n\pi ?

Explanation

Solution

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 ofπ\pi .
A function f(x)f\left( x \right)is said to be periodic if there exists T > 0 such that f (x+T) = f(x)f{\text{ }}\left( {x + T} \right){\text{ }} = {\text{ }}f\left( x \right)for all x in the domain of definition off(x)f\left( x \right). If T is the smallest positive real number such that isf (x+T) = f(x)f{\text{ }}\left( {x + T} \right){\text{ }} = {\text{ }}f\left( x \right), then it is called the period off(x)f\left( x \right).
The trigonometric functions such as sin, cos and tansin,{\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 θ=0sin{\text{ }}\theta = 0other thanθ = 0\theta {\text{ }} = {\text{ }}0.
By seeing the equation, one might straight away reach at the conclusion that θ = 0\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, θ = 0, ± 2π, ± 4π and ± 6π ……… (1)\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’θ = ± π, ± 3π, ± 5π ……… (2)\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 θ = 0sin{\text{ }}\theta {\text{ }} = {\text{ }}0
⇒θ = nπ, where n = 0, ±1, ±2…….\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,
⇒θ = π/2, 5π/2, 9π/2 or, −3π/2, −7π/2   ..……… (1)  \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’
⇒θ = −3π/2, −7π/2 or, −π/2, −5π/2             ………… (2)  \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 ofcos θ = 0 is θ (2n+1) π/2cos{\text{ }}\theta {\text{ }} = {\text{ }}0{\text{ }}is{\text{ }}\theta {\text{ }}\left( {2n + 1} \right){\text{ }}\pi /2, where n = 0, ±1, ±2 ………  n{\text{ }} = {\text{ }}0,{\text{ }} \pm 1,{\text{ }} \pm 2{\text{ }} \ldots \ldots \ldots \;
General solution of the equationsin⁡θ=k\sin \theta = k
We know that whensin⁡θ=k\sin \theta = k, kkhas to be such that −1 ⩽ k ⩽ 1  -1{\text{ }} \leqslant {\text{ }}k{\text{ }} \leqslant {\text{ }}1\;
We can always find some α∈[−π/2, π/2]\alpha \in \left[ {-\pi /2,{\text{ }}\pi /2} \right]
As the function will be: sin (−π)/2 = −1 & sin π/2 = 1sin{\text{ }}\left( { - \pi } \right)/2{\text{ }} = {\text{ }} - 1{\text{ }}\& {\text{ }}sin{\text{ }}\pi /2{\text{ }} = {\text{ }}1, such thatsin⁡θ=k\sin \theta = k, i.e. α = sin−1k\alpha {\text{ }} = {\text{ }}si{n^{ - 1}}k
⇒sin θ = sin α, α∈[−π/2, π/2]\Rightarrow sin{\text{ }}\theta {\text{ }} = {\text{ }}sin{\text{ }}\alpha ,{\text{ }}\alpha \in \left[ {-\pi /2,{\text{ }}\pi /2} \right]
⇒sin θ − sin α = 0\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 ((θ − α)/2) = integral multiple of π\left( {\left( {\theta {\text{ }}-{\text{ }}\alpha } \right)/2} \right){\text{ }} = {\text{ }}integral{\text{ }}multiple{\text{ }}of{\text{ }}\pi
∴θ − α = 2nπ\therefore \theta {\text{ }}-{\text{ }}\alpha {\text{ }} = {\text{ }}2n\pi
Thus θ = 2nπ + α\theta {\text{ }} = {\text{ }}2n\pi {\text{ }} + {\text{ }}\alpha
From equation (1) and (2), we can conclude thatθ = nπ + (−1)n α\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 θ = k  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 θ\theta will be2nπ + α2n\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 preferablysin θ or cos θsin{\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.