Trigonometric Equation | How to solve Trigonometric Equation ?

Trigonometric Equation | How to solve trigonometric equation ?

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

We know that the trigonometric ratio of a certain angle has 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 that, cos30° = √3/2

So, x = 30°

Again, cos30° = cos(360° - 30°) = cos330°

So, x = 330°

∴ x may be 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 solution.

 

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

(ii)     To find the angle in the 1^st 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 2^nd quadrant, we subtract the acute angle x from 180°.

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

(v)    To find the angle in 4^th 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:

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°.

 

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