Question

give the general solution of the following equation cos theta + sin theta is equal to 1
Plz explain also​

Answer 1

AnswEr

The general solution of the given trigonometric equation is :

θ = 2nπ ± π/4 + π/4 where n ∈Z

Given

The equation is

• cosθ+ sinθ = 1

To Find

The value of θ

Solution

cosθ + sinθ = 1

⇒ sin(π/2 - θ) + sinθ = 1

⇒ 2sin[{θ+ π/2 -θ}/2].cos[{θ - (π/2-θ)}/2]=1

⇒2sin(π/4).cos[{θ - π/2 + θ}/2] = 1

⇒ 2×1√2×cos{(2θ - π/2)/2} = 1

⇒ √2cos(θ - π/4) = 1

⇒ cos(θ - π/4) = 1/√2

⇒ cos(θ - π/4) = cos(π/4)

Therefore,

θ - π/4 = 2nπ ± π/4

⇒θ = 2nπ ± π/4 + π/4 where n ∈ Z

So now either

θ = 2nπ + π/4 + π/4

⇒ θ = 2nπ + π/2

Or

θ = 2nπ - π/4 + π/4

⇒ θ = 2nπ

Answer 2

cosθ + sinθ = 1

⇒ sin(π/2 - θ) + sinθ = 1

⇒ 2sin[θ+ π/2 -θ].cos[θ - (π/2-θ)]=1

⇒2sin(π/2).cos[θ - π/2 + θ] = 1

⇒ 2×1×cos(2θ - π/2) = 1

⇒ cos(2θ - π/2) = 1/2

⇒ cos(2θ - π/2) = cos(π/3)

Therefore,

2θ - π/2 = 2nπ ± π/3

⇒2θ = 2nπ ± π/3 + π/2

⇒θ = nπ ± π/6 + π where n ∈ Z

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