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 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.
(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:
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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