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Trigonometric Equation using Formula

We will learn how to solve trigonometric equation using formula.

Here we will use the following formulas to get the solution of the trigonometric equations.

(a) If sin θ = 0 then θ = nπ, where n = 0, ± 1, ± 2, ± 3, …….

(b) If cos θ = 0 then θ = (2n + 1) π2, where n = 0, ± 1, ± 2, ± 3, …….

(c) If cos θ = cos ∝ then θ = 2nπ ± ∝, where n = 0, ± 1, ± 2, ± 3, …….

(d) If sin θ = sin ∝ then θ = n π + (-1) n ∝, where n = 0, ± 1, ± 2, ± 3, …….

(e) If a cos θ + b sin θ = c then θ  = 2nπ + ∝ ±  β, where cos β = ca2+b2, cos ∝ = aa2+b2 and sin ∝ = ba2+b2, where n = 0, ± 1, ± 2, ± 3, …….

1. Solve tan x + sec x = √3. Also find values of x between 0° and 360°.

Solution:

tan x + sec x = √3

sinxcosx + 1cosx = √3, where cos x ≠ 0

⇒ sin x + 1 = √3 cos x

⇒ √3 cos x - sin x = 1,

This trigonometric equation is of the form a cos θ + b sin θ = c where a = √3, b = -1 and c = 1.

⇒ Now dividing both sides by (3)2+(1)2

32 cos x - 12sin x = 12

⇒ cos x cos π4 – sin x sin π6 = cos π3

⇒ cos (x + π6) = cos π3

⇒ x + π6 = 2nπ ± π3, where n = 0, ± 1, ± 2, ± 3, …….

⇒ x = 2nπ ± π3 - π6, where n = 0, ± 1, ± 2, ± 3, …….

When we take minus sign with π3, we get

x = 2nπ - π3 - π6

⇒ x = 2nπ - π2, so that cos x = cos (2nπ - π2) = cos π2 = 0, which spoils the assumption cos x  ≠ 0 (otherwise the given equation would be meaningless).

So, x = 2nπ + π3 - π6, where n = 0, ± 1, ± 2, ± 3, …….

⇒ x = 2nπ + π6, where, n = 0, ± 1, ± 2, ± 3, ……. is the general

solution of the given equation tan x + sec x = √3.

The only solution between 0° and 360° is x = π6 = 30°


2. Find the general solutions of θ which satisfy the equation sec θ = - √2

Solution:   

sec θ = -  √2

⇒ cos θ = - 12

⇒ cos θ = - cos π4

⇒ cos θ = cos (π - π4)

⇒ cos θ = cos 3π4

⇒ θ = 2nπ ± 3π4, where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the general solutions of θ which satisfy the equation sec θ = - √2 is θ = 2nπ ± 3π4, where, n = 0, ± 1, ± 2, ± 3, …….


3. Solve the equation 2 cos2 x + 3 sin x = 0

Solution:

2 cos2 x + 3 sin x = 0

⇒ 2(1 - sin2 x) + 3 sin x = 0

⇒ 2 – 2 sin2 x + 3 sin x = 0

⇒ 2 sin2 x – 3 sin x – 2 = 0

⇒ 2 sin2 x - 4 sin x + sin x – 2 = 0

⇒ 2 sin x(sin x - 2) + 1(sin – 2) = 0

⇒ (sin x - 2)(2 sin x + 1) = 0

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

But sin x – 2 = 0 i.e., sin x = 2, which is not possible.

Now form 2 sin x + 1 = 0 we get

⇒ sin x = -½ 

⇒ sin x =- sin π6

⇒ sin x = sin (π + π6)

⇒ sin x = sin 7π6

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

Therefore, the solution for the equation 2 cos2 x + 3 sin x = 0 is x = nπ + (1)n7π6, where, n = 0, ± 1, ± 2, ± 3, …….

Note: In the above trig equation we observe that there is more than one trigonometric function. So, the identities (sin 2 θ + cos 2 θ = 1) are required to reduce the given equation to a single function.

4. Find the general solutions of cos x + sin x = cos 2x + sin 2x

Solution:

cos x + sin x = cos 2x + sin 2x

⇒cos x - cos 2x - sin 2x + sin x = 0

⇒  (cos x - cos 2x) - (sin 2x - sin x) = 0

⇒  2 sin 3x2 sin x2 - 2 cos 3x2 sin x2 = 0

⇒  sin x2  (sin 3x2 - cos 3x2) = 0
 Therefore, either, sin x2 = 0          

x2= nπ     

⇒ x = 2nπ

or, sin 3x2 -  cos 3x2 = 0

⇒ sin 3x2 = cos 3x2

⇒ tan 3x2 = 1

⇒ tan 3x2 = tan π4

3x2= nπ + π4

⇒ x = 13 (2nπ + π2) = (4n + 1)π6

Therefore, the general solutions of cos x + sin x = cos 2x + sin 2x are x = 2nπ and x = (4n+1)π6, Where, n = 0, ±1, ±2, ………………….. 


5. Find the general solutions of sin 4x cos 2x = cos 5x sin x

Solution:

sin 4x cos 2x = cos 5x sin x

⇒ 2 sin 4x cos 2x = 2 cos 5x sin x

⇒ sin 6x + sin 2x = sin 6x - sin 4x     

⇒ sin 2x + sin 4x =0

⇒ 2sin 3x cos x =0

Therefore, either, sin 3x = 0 or, cos x = 0

i.e., 3x = nπ or, x = (2n + 1)π6

⇒ x = nπ3 or,  x = (2n + 1)π6

Therefore, the general solutions of sin 4x cos 2x = cos 5x sin x are nπ3 and x = (2n + 1)π6

 Trigonometric Equations






11 and 12 Grade Math

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