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# Trigonometric Equations

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#### notes

Equations involving trigonometric functions of a variable are called trigonometric equations.  The solutions of a trigonometric equation for which 0 ≤  x <  2π are called principal solutions.  The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution.
How will solve a trigonometric equation 2"sin"^2 x+ 7"cos"  x= 5? We shall follow few steps:
1) Reduce variables- Try to reduce the equation to one variable.
2(1- cos^2 x)+ 7cos  x= 5

2) Simplify equation
2cos^2 x- 7cos x+ 3= 0
2cos^2 x- 6cos x- cos x+3= 0
2cos x (cos x-3)-1 (cos x- 3)= 0
(2cos x- 1) (cos x- 3)= 0
cos x= 1/2 or cos x= 3

3) Double check domain and range of varibles
The trigonometric function cos lies between -1, 1
therefore, cos x= 1/2

we know that cos 60°= 1/2
thus, x= 60°

4) Find all possible solutions
Principal soultions- 0 ≤ x ≤ 2π
60°, 300°
General solutions-
x= n(2π) ± y
x= 2nπ ± y

#### theorem

Theorem 1: For any real numbers x and y,
sin x = sin y implies x = nπ + (–1)n y, where n ∈ Z
Proof :If sin x = sin y, then
sin x – sin y = 0

or  2cos  (x+y)/2 sin  (x-y)/2= 0

which gives cos   (x+y)/2= 0 or sin  (x-y)/2= 0

Therefore (x+y)/2= (2n+1) π/2 or (x-y)/2= nπ, where n ∈ Z

i.e. x = (2n + 1) π – y  or x = 2nπ + y, where n∈Z
Hence x = (2n + 1)π + (–1)2n + 1 y or x = 2nπ +(–1)2n y, where n ∈ Z.
Combining these two results, we get x = nπ + (–1)n y, where n ∈ Z.

Theorem 2: For any real numbers x and y, cos x = cos y, implies x = 2nπ ± y, where n ∈ Z
Proof: If cos x = cos y, then
cos x – cos y = 0   i.e.,

-2 sin (x+y)/2 sin  (x-y)/2= 0

Thus, sin  (x+y)/2= 0 or sin  (x-y)/2= 0

Therefore, (x+y)/2= nπ  or (x-y)/2= nπ, where n ∈ Z
i.e. x = 2nπ – y  or x = 2nπ + y, where n ∈ Z Hence x = 2nπ ± y, where n ∈ Z

Theorem 3:Prove that if x and y are not odd mulitple of  π/2, then
tan x = tan y implies x = nπ + y, where n ∈ Z
Proof : If tan x = tan y,   then   tan x – tan y = 0
or (sin x cos y- cos x sin y)/(cos x cos y)= 0
which gives sin (x – y) = 0
Therefore x – y = nπ, i.e., x = nπ + y, where n ∈ Z.

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