Question

how to find general solution for theta in radians

Answer 1

Normally, if we are asked to find the solution for θ in the interval (0,2π) for the equation : cos(θ)=12 then we say θ=60∘=π3, but in that interval this is not the only value of θ, θ can have more values like : θ=300∘=5π3..


As all the trigonometric ratios are cyclic or circular in nature, so we will get infinite values for θ that will satisfy the equation cos(θ)=12..


This we can achieve by Finding the General Solution for the given Trigonometric Equation..


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First Form:

1. For sin(θ)=0 , the general solution for θ is : θ=nπ , where n∈Z(Integers)..

Here, n can take integer values like -2, -1, 0, 1 , 2 etc..


2. For cos(θ)=0 , the general solution for θ is : θ=(2n+1)π2, where n∈Z(Integers)..

3. tan(θ)=0 , the general solution for θ is : θ=nπ, where n∈Z(Integers)..

Note: For sin(θ)=0 and tan(θ)=0 , the general solution is same..


Here is an example:


1. For cos(3θ)=0 find the general solution..


Solution:

cos(3θ)=0

⟹3θ=(2n+1)π2 where n∈Z

⟹θ=(2n+1)π6 where n∈Z



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Second Form:

1. For sin(θ)=sin(α), the general solution is : θ=nπ+(−1)nα,n∈Z

2. For cos(θ)=cos(α), the general solution is : θ=2nπ±α,n∈Z

3. For tan(θ)=tan(α), the general solution is : θ=nπ+α,n∈Z..



Here is an example:

For cosec(θ)=2 find the general solution..


Solution:

cosec(θ)=2

⟹1sin(θ)=2 or sin(θ)=12⟹sin(θ)=sin(π6)

So: θ=nπ+(−1)n(π6),n∈Z


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Third Form:

1. For sin2(θ)=sin2(α) and cos2(θ)=cos2(α) and tan2(θ)=tan2(α), the general solution for the three is same and given by:

θ=nπ±α where n∈Z..


Here is an example for this:

For 7cos2(θ)+3sin2(θ)=4 find the general solution..


Solution:

7cos2(θ)+3sin2(θ)=4


7(1−sin2(θ))+3sin2(θ)=4⟹4sin2(θ)=3⟹sin2(θ)=34


⟹sin2(θ)=(3√2)2⟹sin2(θ)=sin2(π3)

⟹θ=nπ±π3 where n∈Z..