The general solution of sinx=1 is given by

x=(4n+1)π/2 where n is any integer.

In this post, we will learn how to find the general solutions of the trigonometric equation sinx=1.

**sinx=1 General Solution**

**Question:** Find the general solutions of sinx=1.

**Answer:**

As sin(π/2)=1, the equation sinx=1 can be written

sinx = sin(π/2).

∴ x=mπ+(-1)^{m} $\frac{\pi}{2}$ for some integers m. [Since the general solutions of sinθ = sinα are given by θ = nπ+(-1)^{n} α, n ∈ Z.]

**Case 1:** First assume that m is even, that is, m=2n where n is an integer. So

x=2nπ+(-1)^{2n} $\frac{\pi}{2}$ = 2nπ+$\frac{\pi}{2}$ = (4n+1)$\frac{\pi}{2}$.

**Case 2:** Next, assume that m is odd, that is, m=2n+1 for some integer n. So

x=(2n+1)π+(-1)^{2n+1} $\frac{\pi}{2}$

= (2n+1)π-$\frac{\pi}{2}$

= 2nπ+$\frac{\pi}{2}$

= (4n+1)$\frac{\pi}{2}$.

Thus, we deduce that the general solution of sinx=1 is equal to x=(4n+1)π/2 where n is any integer. As n can take any integer value, so sinx=1 has infinitely many solutions.

As an application of the above, we can find the general solutions of sin2x=1.

**sin2x=1 General Solution**

**Question:** Find the general solutions of sin2x=1.

**Answer:**

From the above, we know that the general solutions of sinθ=1 are given by θ=(4n+1)π/2 where n ∈ Z.

So the general solutions of sin2x=1 are equal to

2x = (4n+1)π/2 where n ∈ Z.

⇒ x = (4n+1)π/4.

So the general solutions of sin2x=1 are given by x=(4n+1)π/4 where n is an integer.

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## FAQs

**Q1: What are the general solutions of sinx=1?**

**Answer:** The general solutions of sinx=1 are given by x=(4n+1)π/2 where n ∈ Z.

**Q2: What are the general solutions of sinθ=1?**

**Answer:** The general solutions of sinθ=1 are given by θ=(4n+1)π/2 where n is an integer.

**Q3: What are the general solutions of sin2x=1?**

**Answer:** The general solutions of sin2x=1 are given by x=(4n+1)π/4 where n ∈ Z.