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 = \(\frac{π}{2}\).
11 and 12 Grade Math
From sin θ = 1 to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Jun 17, 24 11:22 PM
Jun 16, 24 05:20 PM
Jun 16, 24 04:12 PM
Jun 16, 24 02:34 PM
Jun 16, 24 12:31 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.