Problem with this Trigonometric Equation

Question

I am having trouble figuring out how to solve such an equation can anyone please tell me the steps to solve it as I have been solving a lot of trigonometric equations but I am stuck in this one: $$\sin\left(x+\frac{\pi}{9}\right)=-\frac12$$

Answer 1

Hint: let $x+\frac{\pi}{9}=y$ then you get $$\sin y=-\frac12 \Rightarrow y=-\frac{\pi}{6}+2k\pi \lor y=\frac{7\pi}{6}+2k\pi$$ where $k$ is an integer (because of the periodicity of the function $\sin y$). Finally substitute $x+\frac{\pi}{9}=y$, and you get the solution in terms of $x$.

Answer 2

Take the inverse sine of both sides of the equation. You'll get $$x + \pi/9 = \sin^{-1}(-1/2)$$

Now all you need to do is to determine for what angles $\theta$, $\sin \theta = -\frac 12$.

Answer 3

Notice that $$\sin\theta=-\frac12\iff \theta\equiv-\frac\pi6\mod2\pi\;\;\text{or}\;\;\theta\equiv\frac{7\pi}6\mod2\pi$$

hence in your example we have $$x\equiv-\frac{5\pi}{18}\mod 2\pi\;\;\text{or}\;\;x\equiv\frac{19\pi}{18}\mod2\pi$$