The general solutions of sinx=-1 are 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)$\frac{\pi}{2}$ **…(I) **

**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+$\frac{3\pi}{2}$

= (4n+3)$\frac{\pi}{2}$ **…(II)**

Combining **(I) **and **(II)**, we deduce that x=(4n-1)π/2.

So the general solution of sinx=1 equals x=(4n-1)π/2 where n is any integer. As n can be any integer, so sinx=-1 has infinite solutions.

Let us now find the general solutions of sin2x=-1 as an application of the general solution of sinx=-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.

**ALSO READ:**

tanx=tany General Solution | x=nπ+y; n ∈ Z |

sinx=0 General Solution | x=nπ; n ∈ Z |

sinx=1 General Solution | x=(4n+1)π/2; n ∈ Z |

## 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 is an integer.

**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 ∈ Z.

**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 is an integer.