Question

find the general solution of cosmθ=sinnθ​

Answer 1

Step-by-step explanation:

We have,

cos θ = cos ∝

⇒ cos θ - cos ∝ = 0

⇒ 2 sin (θ+∝)2 sin (θ−∝)2 = 0

Therefore, either, sin (θ+∝)2 = 0 or, sin (θ−∝)2 = 0

Now, from sin (θ+∝)2 = 0 we get, (θ+∝)2 = nπ, n ∈ Z

⇒ θ = 2nπ - ∝, n ∈ Z i.e., (any even multiple of π) - ∝ …………………….(i)

And from sin (θ−∝)2 = 0 we get,

(θ−∝)2 = nπ, n ∈ Z

⇒ θ = 2nπ + ∝, m ∈ Z i.e., (any even multiple of π) + ∝ …………………….(ii)

Now combining the solutions (i) and (ii) we get,

θ = 2nπ ± ∝, where n ∈ Z.

Hence, the general solution of cos θ = cos ∝ is θ = 2nπ ± ∝, where n ∈ Z.

Note: The equation sec θ = sec ∝ is equivalent to cos θ = cos ∝ (since, sec θ = 1cosθ and sec ∝ = 1cos∝). Thus, sec θ = sec ∝ and cos θ = cos ∝ have the same general solution.

Hence, the general solution of sec θ = secs ∝ is θ = 2nπ ± ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)