Trigonometric Equation | How to solve Trigonometric Equation ?

Trigonometric Equation | How to solve trigonometric equation ?

Trigonometric Equation


An equation involving trigonometrical ratios of unknown angles is called a trigonometric equation. The values of the unknown angle which satisfy the given equation are called the solution of the equation.

We know that the trigonometric ratio of a certain angle has a unique value. But, if the value of a trigonometric ratio is given, the angle is not unique. For example, let us take the equation:


2cosx = √3 

or     cosx = √3/2

We know,

√3/2 = cos30°

       cosx = cos30°

i.e.    x = 30°


Again,

cos30° = cos(360° - 30°) = cos330°

i.e.    x = 330°


x maybe 30° or 330°.


Again, if we add 360° to the angle 30° or 330°, cosine of any one of these angles will also be √3/2. So there might be many angles x which satisfy the equation. But, we will try to find the angles within the given range of solutions.


 

Method for finding angles:

(i)      First of all, we determine the quadrant where the angle falls. For this, we use the ‘all, sin, tan, cos’ rule.


If sinx is positive, the angle x falls in the 1st and 2nd quadrants, and if sinx is negative, the angle x falls in the 3rd and 4th quadrants. If cosx is positive, x lies in the 1st and 4th quadrants, and if cosx is negative, x lies in 2nd and 3rd quadrants. If tanx is positive, x lies in the 1st and 3rd quadrants, and if tanx is negative, x lies in the 2nd and 4th quadrants.


4 Quadrants ('All, Sin, Cos,Tan' Rule)

(ii)     To find the angle in the 1st quadrant, we find the acute angle which satisfies the equation.

For example: If 2cosx = 1

Then,  cosx = ½

Or,      cosx = cos60°

So,  x = 60°.


(iii)    To find the angle in 2nd quadrant, we subtract the acute angle x from 180°.


(iv)   To find the angle in 3rd quadrant, we add the acute angle x to 180°.


(v)    To find the angle in 4th quadrant, we subtract the acute angle x from 360°.


(vi)   To find the value of x from the equations like sinx = 0, cosx = 0, tanx = 0, sinx = 1, cosx = 1, sinx = -1 and cosx = -1, we should note the following results:


Value of angles x for trigonometric ratios: sinx = 0, cosx = 0, tanx = 0, sinx = 1, cosx = 1, sinx = -1 and cosx = -1


Worked Out Examples

Example 1: Solve: sin2x = sinx  (0° ≤ x ≤ 180°)


Solution: Here,

sin2x = sinx

or,   sin2x – sinx = 0

or,   sinx(sinx – 1) = 0


Either,  sinx = 0 ………….. (i)

Or,  sinx – 1 = 0 ……..….. (ii)


From (i), sinx = 0 = sin0° = sin180°

x = 0° or 180°


From (ii), sinx – 1 = 0

or, sinx = 1 = sin90°

x = 90°


Hence, x = 0°, 90°, 180°.


 

Example 2: Solve: 7sin2x + 3cos2x = 4  (0° ≤ x ≤ 360°)


Solution: Here,

7sin2x + 3cos2x = 4

or,   7sin2x + 3(1 – sin2x) = 4

or,   7sin2x + 3 – 3sin2x = 4

or,   4sin2x = 4 – 3

or,   4sin2x = 1

or,   sin2x = ¼

or,   sinx = ± ½


Taking +ve sign,

sinx = ½ = sin30°


Again,

sin30° = sin(180° - 30°) = sin150°


x = 30°, 150°


Taking –ve sign,

sinx = - ½ = - sin30° = sin(180° + 30°) = sin210°


Again,

sinx = - ½ = - sin30° = sin(360° - 30°) = 330°


x = 210°, 330°


Hence, x = 30°, 150°, 210°, 330°.

Example 3: Solve: 2cosθ = cotθ  (0° ≤ x ≤ 360°)


Solution: Here,

2cosθ = cotθ

or,   2cosθ = cosθ/sinθ

or,   2sinθ cosθ = cosθ

        [Do not cancel cosθ]

or,   2sinθ cosθ – cosθ = 0

or,   cosθ(2sinθ – 1) = 0


Either,  cosθ = 0 ……….. (i)

Or,  2sinθ – 1 = 0 ……….. (ii)


From (i), cosθ = 0 = cos90° or cos270°

θ = 90° or 270°


From (ii), 2sinθ – 1 = 0

or, 2sinθ = 1

or, sinθ = ½ =sin30°


Again,       

sin30° = sin(180° - 30°) = sin150°

θ = 30°, 150°


Hence, θ = 30°, 90°, 150°, 270°.


 

Example 4: Solve: 2sin2x = 3cosx  (0° ≤ x ≤ 360°)


Solution: Here,

2sin2x = 3cosx

or,   2(1 – cos2x) – 3cosx = 0

or,   2 – 2cos2x – 3cosx = 0

or,   2cos2x + 3cosx – 2 = 0

or,   2cos2x + 4cosx – cosx – 2 = 0

or,   2cosx(cosx + 2) – 1(cosx + 2) = 0

or,   (cosx + 2)(2cosx – 1) = 0


Either, cosx + 2 = 0 ……… (i)

Or, 2cosx – 1 = 0 …………. (ii)


From (i), cosx + 2 = 0

or, cosx = - 2 which is not possible as cosx - 1. So, this equation has no solution.


From (ii), 2cosx – 1 = 0

or, 2cosx = 1

or, cosx = ½ = cos60°


Again,       

cos60° = cos(360° - 60°) = cos300°


x = 60°, 300°.

Example 5: Solve: √3 cosθ + sinθ = 1  (0° ≤ x ≤ 360°)


Solution: Here,

√3 cosθ + sinθ = 1

or,   √3 cosθ = 1 – sinθ

Squaring both sides,

or,   3cos2θ = 1 – 2sinθ + sin2θ

or,   3 – 3sin2θ = 1 – 2sinθ + sin2θ

or,   4sin2θ – 2sinθ – 2 = 0

or,   2sin2θ – sinθ – 1 = 0

or,   2sin2θ – 2sinθ + sinθ – 1 = 0

or,   2sinθ(sinθ – 1) + 1(sinθ – 1) = 0

or,   (sinθ – 1)(2sinθ + 1) = 0


Either, sinθ – 1 = 0 …….. (i)

Or, 2sinθ – 1 = 0 ………... (ii)


From (i), sinθ – 1 = 0

or, sinθ = 1 = sin90°

θ = 90°.


From (ii), 2sinθ + 1 = 0

or, 2sinθ = - 1

or, sinθ = - ½ = - sin30° = sin(180° + 30°) = sin210°


Again,     

- sin30° = sin(360° - 30°) = sin330°

θ = 210°, 330°


Hence, θ = 90°, 210°, 330°.


 

Example 6: Solve: sin3x + sinx = sin2x  (0° ≤ x ≤ 360°)


Solution: Here,

sin3x + sinx = sin2x

or,   2sin[(3x + x)/2] cos[(3x – x)/2] = sin2x

or,   2sin2x cosx – sin2x = 0

or,   sin2x(2cosx – 1) = 0


Either, sin2x = 0 ……….. (i)

Or, 2cosx – 1 = 0 ……….. (ii)


From (i), sin2x = 0 = sin0° or sin180° or sin360°

2x = 0° or 180° or 360°

or, x = 0° or 90° or 180°


From (ii), 2cosx – 1 = 0

or, 2cosx = 1

or, cosx = ½ = cos60°

x = 60°


Hence, x = 0°, 60°, 90°, 180°.

Do you have any questions regarding the trigonometric equation?

 

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