Question

Cosx = rad(3)/2 pls help i dont understand how to solve

Answer 1

Answer:

The general value of x is $x=2n\pi\pm \frac{\pi}{6}$, where n is any integer. The values of x between 0 to 2π are $\frac{\pi}{6}\text{ and }\frac{11\pi}{6}$.

Step-by-step explanation:

It is given that

$\cos x=\frac{\sqrt{3}}{2}$

We know that

$\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$

So, the given equation can be written as

$\cos x=\cos \frac{\pi}{6}$

$x=2n\pi\pm \frac{\pi}{6}$

Where, n is any integer.

For n=0,

$x=2(0)\pi\pm \frac{\pi}{6}=\pm \frac{\pi}{6}$

For n=1,

$x=2(1)\pi\pm \frac{\pi}{6}=2\pi\pm \frac{\pi}{6}=\frac{11\pi}{6},\frac{13\pi}{6}$

Therefore the general value of x is $x=2n\pi\pm \frac{\pi}{6}$, where n is any integer. The values of x between 0 to 2π are $\frac{\pi}{6}\text{ and }\frac{11\pi}{6}$.

Answer 2

Answer
x = 30° or x = (360 – 30) = 330° for value of (0≤x≥360)

Explanation
The first step is to find the anti-cosine of √3/2.
cos^(-1)⁡〖√3/2〗=30°
So, x=30°
The values of x can be many so a limit has to be given. For one complete cycle, the values of x would be.
Since cosx is positive the value of x must have been in the first quadrant and the 4th quadrant.
So, x = 30° or x = (360 – 30) = 330°