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General solution of Trigonometric Equation

We will learn how to find the general solution of trigonometric equation of various forms using the identities and the different properties of trig functions.

For trigonometric equation involving powers, we need to solve the equation either by using quadratic formula or by factoring.

1. Find the general solution of the equation 2 sin3  x - sin x = 1. Hence find the values between 0° and 360° satisfying the given equation.

Solution:

Since the given equation is a quadratic in sin x, we can solve for sin x either by factorization or by using quadratic formula.

Now, 2 sin3  x - sin x = 1

⇒ 2 sin3 x - sin x - 1 = 0

⇒ 2 sin3  x - 2sin x + sin x - 1 = 0

⇒ 2 sin x (sin x - 1) + 1 (sin x - 1) = 0

⇒ (2 sin x + 1)(sin x - 1) = 0

⇒ Either, 2 sin x + 1 = 0 or, sin x - 1 = 0

⇒ sin x = -1/2 or sin x = 1

⇒ sin x = 7π6 or sin x = π2

⇒ x = nπ + (-1)n7π6 or x = nπ + (-1)nπ2, where n = 0, ± 1, ± 2, ± 3, …….

⇒ x = nπ + (-1)n7π6 ⇒ x = …….., π6, 7π6, 11π6, 19π6, ……..  or  x = nπ + (-1)nπ2 ⇒ x = …….., π2, 5π2, ……..

Therefore the solution of the given equation between 0° and 360° are π2, 7π6, 11π6 i.e., 90°, 210°, 330°.

 

2. Solve the trigonometric equation sin3 x + cos3 x = 0 where 0° < x < 360°

Solution:

sin3 x + cos3 x = 0

⇒ tan3 x + 1 = 0, dividing both sides by cos x

⇒ tan3 x + 13 = 0

⇒ (tan x + 1) (tan2 x - tan x + 1) = 0

Therefore, either, tan x + 1 = 0 ………. (i) or, tan2 x - tan θ + 1 = 0 ………. (ii)

From (i) we get,

tan x = -1

⇒ tan x = tan (-π4)

⇒ x = nπ - π4

From (ii) we get,

tan2 x - tan θ + 1 = 0

⇒ tan x = 1±141121

⇒ tan x = 1±32

Clearly, the value of tan x, are imaginary; hence, there is no real solution of x

Therefore, the required general solution of the given equation is:

x = nπ - π4 …………. (iii) where, n = 0, ±1, ±2, …………………. 

Now, putting n = 0 in (iii) we get, x = - 45°

Now, putting n = 1 in (iii) we get, x = π - π4 = 135°

Now, putting n = 2 in (iii) we get, x = π - π4 = 135°

Therefore, the solutions of the equation sin3  x + cos3  x = 0 in 0° < θ < 360° are x = 135°, 315°.


3. Solve the equation tan2 x = 1/3 where, -  π ≤ x ≤ π.

 Solution: 

tan 2x= 13

⇒ tan x= ± 13

⇒ tan x = tan (±π6

Therefore, x= nπ  ±  π6,  where  n = 0, ±1, ±2,………… 

When, n = 0 then x = ± π6 = π6 or,- π6

If n = 1 then x = π ± π6 + 5π6 or,- 7π6

If n = -1 then x = - π ± π6 =- 7π6, - 5π6

Therefore, the required solutions in – π ≤ x ≤ π are x = π6, 5π6, - π6, - 5π6.

 Trigonometric Equations






11 and 12 Grade Math

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