## Note on Solution of Trigonometric Equations

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Equalities like 1 - sin2θ = cos2θ , 1 + cos2θ = 2 cos2θ etc are satisfied by all the values of the angle θ . So, they are called identities.

Let the equalities like 2cosθ = 1,$$\sqrt 2$$ sinθ = 1 etc. are satisfied by only some values of the angle θ.

A trigonometric ratio of a certain angle has one and only value. But, if the value of a trigonometric ratio is given, the angle is not unique. For example, let us take the equation

2cosx = $$\sqrt 3$$ or cos x = $$\frac{\sqrt 3}{2}$$

We know that cos300 = $$\frac{\sqrt 3}{2}$$

So, x = 300

Again, cos3300= cos(3600 - 300) = cos300 = $$\frac{\sqrt 3}{2}$$

So, x = 3300

∴ x may be 300 or 3300.

Again if we add 3600 or multiple of 3600 to the angle 300 or 3300, the cosine of any one of these angles will also be $$\frac{\sqrt 3}{2}$$. So there might be many values of x which satisfy the equation. But we will try to find those angles which lie between 00 and 3600.

A 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 sinθ is positive, the angle θ falls in the 1st and 2nd quadrants and if sinθ is negative, the angle θ falls in the 3rd and 4th quadrants. If cosθ is positive, θ lies in the 1st and 4th quadrants and if cosθ is negative, θ lies in the 2nd and 3rd quadrants. If tanθ is positive, θ lies in the first and third quadrants and if tanθ is negative, θ lies in second and fourth quadrants.

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

For example, if 2cosθ = 1

then cosθ = $$\frac{1}{2}$$

or, cosθ = cos600

So, θ = 600

(iii) To find the angle in the second quadrant, we subtract acute angle θ from 1800.

(iv) To find the angle in the third quadrant, we add acute angle θ to 1800

(v) To find the angle in the fourth quadrant, we subtract acute angle θ to 3600

(vi) To find the value of θ from the equations like sinθ = 0, cosθ = 0, tanθ= 0, sinθ = 1, cosθ = 1, sinθ = -1, cosθ = -1. We should note the following results:

 If sinθ = 0, thenθ = 00, 1800 or 3600 If sinθ = 1, thenθ = 900 If tanθ = 0, thenθ = 00,1800 or 3600 If sinθ = -1, thenθ = 2700 If cosθ = 0, thenθ = 900 or 2700 If cosθ = 1, thenθ = 00 or 3600 If cosθ = -1, then θ = 1800.

A 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.

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

(iii) To find the angle in the second quadrant, we subtract acute angle θ from 1800.

(iv) To find the angle in the third quadrant, we add acute angle θ to 1800

(v) To find the angle in the fourth quadrant, we subtract acute angle θ to 3600

.

### Very Short Questions

Here,

2 cos$$\theta$$ = 1

or, cos$$\theta$$ = $$\frac 12$$

or, cos$$\theta$$ = cos 60° , cos (360° - 60°)

∴ $$\theta$$ = 60° , 300° Ans

Here,

3 tan$$\theta$$ = $$\sqrt 3$$

or, tan$$\theta$$ = $$\frac {\sqrt 3}3$$

or, tan$$\theta$$ = $$\frac {\sqrt 3}{\sqrt 3 × \sqrt 3}$$

or, tan$$\theta$$ = $$\frac 1{\sqrt 3}$$

or, tan$$\theta$$ = tan 30° , tan (180° + 30°)

∴ $$\theta$$ = 30° , 210° Ans

Here,

$$\sqrt 2$$ sec$$\theta$$ + 2 = 0

or, $$\sqrt 2$$ sec$$\theta$$ = - 2

or. sec$$\theta$$ = $$\frac {-2}{\sqrt 2}$$

or, sec$$\theta$$ = -$$\frac {\sqrt 2 × \sqrt 2}{\sqrt 2}$$

or, sec$$\theta$$ = -$$\sqrt 2$$

or, sec $$\theta$$ = sec (180° - 45°), sec (180v + 45°)

∴ $$\theta$$ = 135° , 225° Ans

Here,

tanx - sinx = 0

or, $$\frac {sinx}{cosx}$$ - sinx = 0

or, $$\frac {sinx - sinx cosx}{cosx}$$ = 0

or, sinx (1 - cosx) = 0

Either,

sinx = 0

sinx = sin 0°

∴ x = 0°

Or,

1 - cosx = 0

or, cosx= 1

or, cosx = cos180°

∴ x = 180°

∴ x = 0° , 180° Ans

Here,

2 sin$$\theta$$ + 1 = 0

or, 2 sin$$\theta$$ = - 1

or, sin$$\theta$$ = -$$\frac 12$$

or, sin$$\theta$$ = sin (180° + 30°), sin (360° - 30°)

∴ $$\theta$$ = 210°, 330° Ans

Here,

cosx + secx = $$\frac 52$$

or, cosx + $$\frac 1{cosx}$$ = $$\frac 52$$

or, $$\frac {cos^2x + 1}{cosx}$$ = $$\frac 52$$

or, 2 cos2x + 2 = 5 cosx

or, 2 cos2x - 5 cosx+ 2 = 0

or, 2 cos2x - 4 cosx - cosx + 2 = 0

or, 2 cosx (cosx - 2) - 1 (cosx - 2) = 0

or, (cosx - 2) (2 cosx - 1) = 0

Either,

cosx - 2 = 0

or, cosx = 2 (Impossible)

Or,

2 cosx - 1 = 0

or, 2 cosx = 1

or, cosx = $$\frac 12$$

or, cosx = cos 60°

∴ x = 60° Ans

Here,

4 sin$$\alpha$$ = 3 cosec$$\alpha$$

or, 4 sin$$\alpha$$ = 3$$\frac 1{sin\alpha}$$

or, 4 sin2$$\alpha$$ = 3

or, sin2$$\alpha$$ = $$\frac 34$$

∴ sin$$\alpha$$ =± $$\frac {\sqrt 3}2$$

Taking positive,

sin$$\alpha$$ = $$\frac {\sqrt 3}2$$

or, sin$$\alpha$$ = sin 60° , sin (180° - 60°)

∴ $$\alpha$$ = 60° , 120°

Taking negative,

sin$$\alpha$$ = -$$\frac {\sqrt 3}2$$

or, sin$$\alpha$$ = sin (180° + 60°), sin (360° - 60°)

∴ $$\alpha$$ = 240° , 300°

∴ $$\alpha$$ = 60°, 120°, 240°, 300° Ans

Here,

cos2$$\theta$$ - 1 = sin$$\theta$$

or, 1 - sin2$$\theta$$ - 1 - sin$$\theta$$ = 0

or, - sin2$$\theta$$ - sin$$\theta$$ = 0

or, - sin$$\theta$$ (sin$$\theta$$ - 1) = 0

Either,

sin$$\theta$$ = 0

or, sin$$\theta$$ = sin 0°, sin 180°

Or,

sin$$\theta$$ - 1 = 0

or, sin$$\theta$$ = 1

or, sin$$\theta$$ = sin 90°

∴ $$\theta$$ = 0° , 90° , 180° Ans

Here,

sin$$\theta$$ - cos$$\theta$$ = 0

or, sin$$\theta$$ = cos$$\theta$$

or, $$\frac {sin\theta}{cos\theta}$$ = 1

or, tan$$\theta$$ = tan 45°

∴ $$\theta$$ = 45° Ans

Here,

cot2x + cosec2x = 3

or, cosec2x - 1 + cosec2x = 3

or, 2 cosec2x = 3 + 1

or, cosec2x = $$\frac 42$$

or, cosec2x = 2

or, cosec x =± $$\sqrt 2$$

For positive sign,

cosec x = cosec 45° , cosec (360° - 45°)

For negative sign,

cosec x = cosec (180° - 45°) , cosec (180° + 45°)

∴ x = 45°, 135°, 225° and 315° Ans

Here,

cos 2x = sin x

or, cos 2x = cos (90° - x)

or, 2x = 90° - x

or, 3x = 90°

or, x = $$\frac {90°}3$$

∴ x = 30° Ans

Here,

sin$$\theta$$ + cos$$\theta$$ = $$\sqrt 2$$

Squaring on both sides:

(sin$$\theta$$ + cos$$\theta$$)2= ($$\sqrt 2$$)2

or, sin2$$\theta$$ + cos2$$\theta$$ + 2 sin$$\theta$$ cos$$\theta$$ = 2

or, 1 + sin 2$$\theta$$ = 2

or, sin 2$$\theta$$ = 2 - 1

or, sin 2$$\theta$$ = 1

or, sin 2$$\theta$$ = sin 90°

or, 2$$\theta$$ = 90°

or, $$\theta$$ = $$\frac {90°}2$$

∴ $$\theta$$ = 45° Ans

Here,

2 cos2$$\theta$$ = - $$\sqrt 3$$ cos$$\theta$$

or, 2 cos2$$\theta$$ + $$\sqrt 3$$ cos$$\theta$$ = 0

or, cos$$\theta$$ (2 cos$$\theta$$ + $$\sqrt 3$$) = 0

Either,

cos$$\theta$$ = 0

or, cos$$\theta$$ = cos 90°

Or,

2 cos$$\theta$$ + $$\sqrt 3$$ = 0

or, 2 cos$$\theta$$ = - $$\sqrt 3$$

or, cos$$\theta$$ = - $$\frac {\sqrt 3}2$$

or, cos$$\theta$$ = cos (180° - 30°)

∴ $$\theta$$ = 90° and 150° Ans

Here,

cos2$$\frac {\theta}2$$ - cos$$\frac {\theta}2$$ + $$\frac 14$$ = 0

or, (cos$$\frac {\theta}2$$)2 - 2 . cos$$\frac {\theta}2$$ . ($$\frac 12$$)2 = 0

or, (cos$$\frac {\theta}2$$ - $$\frac 12$$)2 = 0

or, cos$$\frac {\theta}2$$ - $$\frac 12$$ = 0

or, cos$$\frac {\theta}2$$ = $$\frac 12$$

or, cos$$\frac {\theta}2$$ = cos 60°

or, $$\frac {\theta}2$$ = 60°

or, $$\theta$$ = 2× 60°

∴ $$\theta$$ = 120° Ans

0%

40(^o)

30(^o)

50(^o)

60(^o)

40(^o)

60(^o)

30(^o)

50(^o)

• ### 2cos2θ - 1 = 0.

50(^o) , -135(^o)

45(^o) , 135(^o)

-45(^o) , -135(^o)

35(^o) , 35(^o)

0o , 180o

100o , 100o

0o , -180o

360o , 80o

155(^o)

135(^o)

-35(^o)

360(^o)

-60(^o)

60(^o)

10(^o)

30(^o)

90(^0)

30(^0)

-30(^0)

60(^0)

- 45(^0)

45(^0)

35(^0)

55(^0)

30(^0)

60(^0)

90(^0)

120(^0)

90(^0)

60(^0)

260(^0)

180(^0)

• ### Sollve: 2 cos2 ( heta) = 3 sin ( heta)    (    00 ≤ ( heta) ≤ 1800     )

50(^0) , 190(^0)

40(^0) , 180(^0)

30(^0) , 150(^0)

-80(^0) , 90(^0)

• ### Solve:2cos2 ( heta) - cos ( heta) = 2                                       [00≤ ( heta)≤ 1800]

10(^o) ,39(^o) ,120(^o) , 180(^o)

0(^o) ,60(^o) ,150(^o) , 190(^o)

0(^o) ,130(^o) ,150(^o) , 360(^o)

0(^o) ,30(^o) ,150(^o) , 180(^o)

• ### 2sin2( heta) - cos( heta) = 1 .

60(^o) ,180(^o) , 300(^o)

120(^o) ,180(^o) , 300(^o)

60(^o) ,160(^o) , 360(^o)

120(^o) ,190(^o) , 310(^o)

• ### 7sin2 ( heta) + 3cos2 ( heta) = 4

-1

0(^o) , cos (^{-1}) (13) , 360(^o)

0(^o) , cos (^{1}) (-13) , 36(^o)

1

• ### 3sin2x + 4cosx = 4

0o , 80.5o , 219.50 , 360o

0o , 70.5o , 289.50 , 360o

0o , 70o , 2890 , 160o

-1

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tan105

##### sahil

tanθ tan2θ √3tanθ.tan2θ=√3

##### I have qns on trigonometry...

The qns is to prove:- 16cos24.cos48.cos96.cos168=1