# FIND ALL SOLUTIONS OF THE EQUATION EXPRESS THE SOLUTIONS IN RADIANS

## About "SOLVING EQUATIONS IN THE FORM a Sin Theta Plus b Cos Theta Equals c"

Solving Equations in the Form a Sin Theta Plus b Cos Theta Equals c :

Here we are going to see how to solve trigonometric equations in the form a sin θ + b cos θ = c

 Trigonometric equationsin θ = 0cos θ = 0tan θ = 0sin θ = sinα, where α ∈ [−π/2, π/2]cos θ = cos α, where α ∈ [0,π]tan θ = tanα, where α ∈ (−π/2, π/2) General solutionθ = nπ; n ∈ Zθ = (2n + 1) π/2; n ∈ Zθ = nπ; n ∈ Zθ = nπ + (−1)n α, n ∈ Zθ = 2nπ ± α, n ∈ Zθ = nπ + α, n ∈ Z

## Solving Equations in the Form a Sin Theta Plus b Cos Theta Equals c - Examples

Question 1 :

Solve the following equation

(vi)  sin θ + cos θ = 2

Solution :

sin θ + cos θ = √2

Divide by 2 on both sides

(1/2) sin θ + (1/2) cos θ = 1

cos (π/4) cos θ - sin (π/4) sin θ = cos 0

cos (π/4 - θ)  =  cos 0

θ = 2nπ + a

θ = 2nπ + π/4

θ = (8n + 1)π/4

(vii) sin θ + √3 cos θ = 1

Solution :

r = √a2 + b2

a = 1, b = √3, then r = √12 + √3 =  2

Divide the given equation by "r", we get

(1/2) sin θ + √3/2 cos θ = 1/2

sin(π/6) sin θ + cos(π/6) cos θ = 1/2

cos (π/6 - θ)  =  cos-1(1/2)

cos (θ-π/6)  =  cos-1(1/2)

θ  =  2nπ ± π/3

θ - π/6  =  2nπ ± π/3

θ  =  2nπ ± π/3 + π/6

(viii) cot θ + cosecθ  = 3

Solution :

cot θ + cosecθ  = √3

(cos θ /sin θ) + (1/sin θ)  =  √3

(cos θ + 1)/sin θ  =  √3

cos θ + 1 √3 sin θ

√3 sin θ - cos θ  = 1  ----(10)

a = √3, b = -1

r =  √a2 + b2   √(√3)2 + (-1)2   √4  =  2

Divide (1) by 2, we get

(√3/2) sin θ - (1/2) cos θ  = (1/2)

(1/2) cos θ - (√3/2) sin θ = (-1/2)

cos (π/3) cos θ - cos (π/3) sin θ = cos 2π/3

cos ((π/3) + θ)  =  cos 2π/3

θ  =  2nπ ± a

(π/3) + θ  =  2nπ ± 2π/3

θ  =  2nπ ± (2π/3) - (π/3)

(ix)  tan θ + tan (θ + π/3) + tan (θ + 2π/3)  =  3

Solution :

tan (θ + π/3)  =  (taθ + tan π/3) / (1 - taθ tan π/3)

tan (θ + π/3)  =  (taθ + 3/ (1 - taθ)  ------(1)

tan (θ + 2π/3)  =  (taθ + tan 2π/3) / (1 - taθ tan 2π/3)

tan (θ + 2π/3)  =  (taθ - √3) / (1 + √3 taθ) ------(2)

(1) + (2)

=  (taθ+3/ (1-taθ) + (taθ-√3) / (1+√3taθ)

=  [(taθ+3)(1+√3taθ) + (1-taθ) (taθ-√3)]/(1 - 3tanθ)

=  8 tan θ / (1 - 3 tanθ)

tan θ + (1) + (2)  =  √3

tan θ + (8 tan θ / (1 - 3 tanθ))  =  √3

(tan θ - 3 tanθ + 8 tan θ )/(1 - 3 tanθ)  =  √3

(9 tan θ - 3 tanθ)/(1 - 3 tanθ)  =  √3

3 (3 tan θ - tanθ)/(1 - 3 tanθ)  =  √3

tan 3θ  =  √3

tan 3θ  =  √3/3

tan 3θ  =  1/√3

For tan θ, we have θ = nπ + a

3θ = nπ + π/6

θ = nπ/3 + π/18

(x)  cos 2θ = (√5 + 1)/4

Solution :

cos 2θ = (√5 + 1)/4

cos 2θ  =  cos 36

θ = 2nπ ± a

2θ = 2nπ ± 36

θ = nπ ± 18

θ = nπ ± π/10

(xi)  2 cos2 x − 7 cos x + 3 = 0

Solution :

2 cos2 x − 7 cos x + 3 = 0

Let t = cos x

2 t2 − 7t + 3 = 0

(2t - 1)(t - 3)  = 0

 2t - 1 = 02t  =  1t = 1/2cos x = 1/2 t - 3 = 0t = 3

θ = 2nπ ± a

a = π/3

θ = 2nπ ± π/3 After having gone through the stuff given above, we hope that the students would have understood, "Solving Equations in the Form a Sin Theta Plus b Cos Theta Equals c"

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