Trigonometric Equations and its Solutions

Trigonometric equations are those which involve trigonometric ratios. Finding the solution of a trigonometric equation requires to find the value of the angles satisfying the equation. If the interval of the solution is provided, then the value of angles has to be found out. If the interval is not given, then the general solution has to be found.

What is System Of Trigonometric Equations?

The mathematical expressions which include the ratios of the sides of a right-angled triangle can be called trigonometric expressions. An algebraic equation, which involves trigonometric ratios and expressions is known as a trigonometric equation.

The solutions are principal solutions if the variable of the equation lies in the interval 0 and 2π. It comprises all the solutions of trigonometric equations.

Important Results

The following formulae are used when solving a trigonometric equation:

Trigonometrical equation General Solution
sin θ = 0 Then θ = nπ
cos θ = 0 θ = (nπ + π/2)
tan θ = 0 θ = nπ
sin θ = 1 θ = (2nπ + π/2) = (4n+1)π/2
cos θ = 1 θ = 2nπ
sin θ = sin α θ = nπ + (-1)nα, where α ∈ [-π/2, π/2]
cos θ = cos α θ = 2nπ ± α, where α ∈ (0, π]
tan θ = tan α θ = nπ + α, where α ∈ (-π/2 , π/2]
sin 2θ = sin 2α θ = nπ ± α
cos 2θ = cos 2α θ = nπ ± α
tan 2θ = tan 2α θ = nπ ± α

Procedure To Find The General Solution Of Trigonometric Equations

1] Deduce the given equation in the form of sinx, cosx, tanx.

2] Transform them into below forms

sinx = sin y

cos x = cos y

tan x = tan y

Use the first principle value of x as y. 

3] To find the general solutions of trigonometric equations the formulae are used.

sinx=siny then x=nπ+(1)ny where n is any integercosx=cosy then x=2nπ+yorx=2nπy where n is any integertanx=tany then x=nπ+y where n is any integersinx =siny \text \ then \ x=n \pi + (-1)^{n}y \text \ where \ n \ is \ any \ integer\\ cosx=cosy \text \ then \ x=2n \pi + y or x=2n \pi – y \text \ where \ n \ is \ any \ integer\\ tanx=tany \text \ then \ x=n \pi +y \text \ where \ n \ is \ any \ integer\\

Also read

Trigonometric equations

Solved Examples for IIT JEE

Example 1: If sinθ + cosθ = 1 then what is the general value of θ?

Solution:

sinθ + cosθ = 1 ⇒ (1 / √2) sinθ + (1 / √2) cosθ = 1 / √2

dividing by √[12 + 12] = √2,

we get sin (θ + π / 4) = [1/ √2] = sin π / 4

⇒ θ + π / 4 = nπ + (−1)n [π / 4]

⇒ θ = nπ + (−1)n [π / 4] − π / 4

Example 2: If tan (cot x) = cot (tan x) , then what is sin 2x?

Solution:

Trigonometric Equation Examples

Example 3: If the two angles on the base of a triangle are (22.5)o and (112.5)o, then what is the ratio of the height of the triangle to the length of the base?

Solution:

In Δ ACD, h / sin (67.5o) = AC / sin 90o

⇒ h / AC = sin (67.5o) ….(i)

In ΔABC, AC / sin (22.5o) = x / sin 45o

⇒ AC / x = √2 / sin (22.5o) …..(ii)

From (i) and (ii),

h / x = 1 / 2

Example 4: If the angles of a triangle be in the ratio 1 : 2 : 7, then what is the ratio of its greatest side to the least side?

Solution:

x + 2x + 7x = 180o⇒ x = 18o

Hence the angles are 18o, 36o, 126o

Greatest side = sin (126o)

Smallest side = sin (18o) and

Ratio = sin 126o / sin (18o) = [√5 + 1] / [√ 5 − 1]

Example 5: If the angles A, B, C of a triangle are in A.P. and the sides a, b, c opposite to these angles are in G. P. then a2, b2, c2 are in which progression?

Solution:

Since A, B and C are in A.P., therefore B = 60 and b2 = ac

cos B = a2 + c2−b2 / 2ac

⇒ 1 / 2 = a2 + c2−b2 / 2b2 (∵ b2 = ac)

b2 = a2 + c2−b2

⇒ a2 + c2 = 2b2

Example 6: Find the Principal solutions of the below trigonometric equations

a. sin x = 1/2

b. cos x  = -1/2

c. tan x = -1

Solution: The principle solution is in the range 0 ≤ x ≤ 2π. 

To find the value of the angle using the positive value.

The following identities need to be recalled to find the principal values.

sin(πx)=sinxsin(π+x)=sinxsin(2πx)=sinxcos(πx)=cosxcos(π+x)=cosxcos(2πx)=cosxtan(πx)=tanxtan(π+x)=tanxtan(2πx)=tanxsin( \pi -x )=sin x\\ sin ( \pi + x) = -sinx \\ sin ( 2\pi – x) = -sinx \\ cos ( \pi -x )= -cos x\\ cos ( \pi +x )= -cos x\\ cos ( 2\pi -x )= cos x\\ tan ( \pi -x )= -tan x\\ tan ( \pi +x )= tan x\\ tan ( 2\pi -x )= -tan x\\

a) sin x = 1/2

sinx=sin(π/6)sin(ππ/6)=sinπ/6sin(5π/6)=1/2sin x = sin (\pi/6)\\ sin (\pi – \pi/6) = sin \pi/6\\ sin (5 \pi/6) = 1/2\\

The principle solutions are π/6,  5π/6.

 b) cos x = -1/2

cosx=cos(π/3)cos(ππ/3)=cos(π/3)cos(2π/3)=1/2cos(π+π/3)=cos(π/3)cos(4π/3)=1/2cos x = -cos (\pi /3)\\ cos (\pi – \pi /3) = -cos (\pi /3)\\ cos (2 \pi/3) = -1/2\\ cos (\pi + \pi /3)= -cos (\pi /3)\\ cos (4 \pi/3) = -1/2\\

The principle solutions are 2π/3, 4π/3.

 c) tanx =-1

tanx=tan(π/4)tan(ππ/4)=tan(π/4)tan(3π/4)=1tan(2ππ/4)=tan(π/4)tan(7π/4)=1tan x = -tan (\pi/4)\\ tan ( \pi -\pi/4 )= -tan (\pi/4)\\ tan (3\pi/4) = -1\\ tan ( 2\pi – \pi/4 )= -tan (\pi/4)\\ tan (7 \pi/4) = -1\\

The principle solutions are 3π/4, 7π/4.

Example 7: Find the general solutions of the following.

a. sinx=12b. cosx=32c. cosx=12a. \ sin x = – \frac {1}{2}\\ b. \ cos x = \frac {\sqrt {3}}{2}\\ c. \ cos x = \frac {1}{2}\\

Solution: 

Using the steps described above, 

a. sinx=12sinx=sin(π+π/6)=sin(7π/6)x=nπ+(1)n7π6 where n is any integer is the general solutiona. \ sin x = – \frac {1}{2}\\ sin x = sin (\pi + \pi/6) = sin (7 \pi/6)\\ x=n \pi + (-1)^{n} \frac {7 \pi}{6} \text \ where \ n \ is \ any \ integer \ is \ the \ general \ solution\\

b. cosx=32cosx=cos(π/6)x=2nπ+π6 orx=2nππ6 where n is any integer is the general solutionb. \ cos x = \frac {\sqrt {3}}{2}\\ cos x = cos (\pi/6)\\ x=2n \pi + \frac {\pi}{6} \text \ or x=2n \pi – \frac {\pi}{6} \text \ where \ n \ is \ any \ integer \ is \ the \ general \ solution\\

c. cosx=12cosx=cos(π/3)x=2nπ+π3 orx=2nππ3 where n is any integer is the general solutionc.\ cos x = \frac {1}{2}\\ cos x = cos (\pi/3)\\ x=2n \pi + \frac {\pi}{3} \text \ or x=2n \pi – \frac {\pi}{3} \text \ where \ n \ is \ any \ integer \ is \ the \ general \ solution\\

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