Question

2cosx+sqrt3=0 sin(2P-x)-cos(3P/2+x)+1=0 sin9P/4 cos(-4P/3)

Answer 1

$2cosx+\sqrt3=0\\cosx=-\frac{\sqrt3}{2}\\x=\pm arccos(-\frac{\sqrt3}{2})+2\pi n\\x=\pm\frac{\pi}{6}+2\pi n, \; n\in Z;\\\\sin(2\pi-x)-cos(\frac{3\pi}{2}+x)+1=0\\-sinx-sinx+1=0\\-2sinx=-1\\sinx=\frac{1}{2}\\x=(-1)^narcsin\frac{1}{2}+\pi n\\x=(-1)^n\frac{\pi}{6}+\pi n, \; n\in Z;\\\\sin\frac{9\pi}{4}=sin(\frac{8\pi}{4}+\frac{\pi}{4})=sin(2\pi+\frac{\pi}{4})=sin\frac{\pi}{4}=\frac{\sqrt2}{2};\\\\cos(-\frac{4\pi}{3})=cos(-(\frac{3\pi}{3}+\frac{\pi}{3}))=cos(\pi+\frac{\pi}{3})=-cos\frac{\pi}{3}=-\frac{1}{2}.$

Answer 2

It seems like you've provided a set of trigonometric expressions. Let's work through them one by one:

1. $2\cos(x) + \sqrt{3} = 0$:

To solve for $x$, we can isolate $\cos(x)$:

$2\cos(x) = -\sqrt{3}$

Now, divide both sides by 2:

$\cos(x) = -\frac{\sqrt{3}}{2}$

This equation corresponds to the angle $x$ where the cosine function equals $-\frac{\sqrt{3}}{2}$, which occurs in the second and third quadrants for the unit circle. Therefore, the solutions are:

$x = \frac{5\pi}{6} + 2n\pi$ and $x = \frac{7\pi}{6} + 2n\pi$, where $n$ is an integer.

2. $\sin(2\pi - x) - \cos\left(\frac{3\pi}{2} + x\right) + 1 = 0$:

First, let's simplify the angles:

$\sin(2\pi - x) = \sin(-x)$ (since $\sin(2\pi) = 0$).

$\cos\left(\frac{3\pi}{2} + x\right) = -\sin\left(\frac{\pi}{2} - x\right)$ (using the complementary angle identity).

Now, the equation becomes:

$\sin(-x) + \sin\left(\frac{\pi}{2} - x\right) + 1 = 0$

We can use the sum-to-product trigonometric identity $\sin(A) + \sin(B) = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)$:

$2\sin\left(\frac{-x + \frac{\pi}{2} - x}{2}\right)\cos\left(\frac{-x - \frac{\pi}{2} + x}{2}\right) + 1 = 0$

Simplify further:

$2\sin\left(\frac{-2x + \frac{\pi}{2}}{2}\right)\cos\left(\frac{-\frac{\pi}{2}}{2}\right) + 1 = 0$

$\sin\left(-x + \frac{\pi}{4}\right)\cos\left(\frac{-\pi}{4}\right) + 1 = 0$

Now, we have:

$\sin\left(-x + \frac{\pi}{4}\right) = -\sqrt{2}\sin\left(-x + \frac{\pi}{4}\right)$

So, the equation becomes:

$-\sqrt{2}\sin\left(-x + \frac{\pi}{4}\right) - \frac{1}{\sqrt{2}} + 1 = 0$

Now, solve for $-\sqrt{2}\sin\left(-x + \frac{\pi}{4}\right)$