sin θ = 1

How to find the general solution of an equation of the form sin θ = 1?

Prove that the general solution of sin θ = 1 is given by θ = (4n + 1)π/2, n ∈ Z.

Solution:

We have,

sin θ = 1       

⇒ sin θ = sin \(\frac{π}{2}\)

θ = mπ + (-1)\(^{m}\) ∙ \(\frac{π}{2}\), m ∈ Z, [Since, the general solution of sin θ = sin ∝ is given by θ = nπ + (-1)\(^{n}\) ∝, n ∈ Z.]

Now, if m is an even integer i.e., m = 2n (where n ∈ Z) then,

    θ = 2nπ + \(\frac{π}{2}\)

⇒ θ = (4n + 1)\(\frac{π}{2}\)

Again, if m is an odd integer i.e. m = 2n + 1 (where n ∈ Z) then,

θ = (2n + 1) ∙ π - \(\frac{π}{2}\)

⇒ θ = (4n + 1)\(\frac{π}{2}\).

Hence, the general solution of sin θ = 1 is θ = (4n + 1)\(\frac{π}{2}\), n ∈ Z.


1. Solve the trigonometric equation sin x - 2 = cos 2x, (0 ≤ x ≤ \(\frac{π}{2}\))

Solution:

sin x - 2 = cos 2x

⇒ sin x - 2 = 1 - 2 sin 2x

⇒ 2 sin\(^{2}\) x + sin x - 3 = 0

⇒ 2 sin\(^{2}\) x + 3 sin x - 2 sin x - 3 = 0

⇒ sin x (2 sin x + 3) - 1(2 sin x + 3) = 0

⇒ (2 sin x + 3) (sin x - 1) = 0

Therefore, either, 2 sin x + 3 = 0 ⇒ sin x = - \(\frac{3}{2}\), Which is impossible since the numerical value of sin x cannot be greater than 1.

or, sin x - 1 = 0 

⇒ sin x = 1

We know that the general solution of sin θ = 1 is θ = (4n + 1)\(\frac{π}{2}\), n ∈ Z.

Therefore, x = (4n + 1)\(\frac{π}{2}\) …………… (1) where, n ∈ Z.

Now, Putting n = 0 in (1) we get, x = \(\frac{π}{2}\)

Now, Putting   n = 1 in (1) we get, x = \(\frac{5π}{2}\)

Therefore, the required solution in 0 ≤ x ≤ 2π is:   x = \(\frac{π}{2}\).

 Trigonometric Equations






11 and 12 Grade Math

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