**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*

*Worked Out Examples*

*Example 1: Solve: sin ^{2}x = sinx (0° *

*≤ x ≤ 180°)**Solution: **Here,*

*sin ^{2}x = sinx*

*or, sin ^{2}x
– 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: 7sin ^{2}x + 3cos^{2}x
= 4 *

*(0°*

*≤ x ≤ 360°)**Solution: **Here,*

*7sin ^{2}x + 3cos^{2}x = 4*

*or, 7sin ^{2}x
+ 3(1 – sin^{2}x) = 4*

*or, 7sin ^{2}x
+ 3 – 3sin^{2}x = 4*

*or, 4sin ^{2}x
= 4 – 3*

*or, 4sin ^{2}x
= 1*

*or, sin ^{2}x
= 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: 2sin ^{2}x = 3cosx *

*(0°*

*≤ x ≤ 360°)**Solution: **Here,*

*2sin ^{2}x = 3cosx*

*or, 2(1
– cos ^{2}x) – 3cosx = 0*

*or, 2 –
2cos ^{2}x – 3cosx = 0*

*or, 2cos ^{2}x
+ 3cosx – 2 = 0*

*or, 2cos ^{2}x
+ 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, 3cos ^{2}θ
= 1 – 2sinθ + sin^{2}θ*

*or, 3 –
3sin ^{2}θ = 1 – 2sinθ + sin^{2}θ*

*or, 4sin ^{2}θ
– 2sinθ – 2 = 0*

*or, 2sin ^{2}θ
– sinθ – 1 = 0*

*or, 2sin ^{2}θ
– 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°.*

*You can comment your
questions or problems regarding the solution of trigonometric equations here.*

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