1. Trig Equations

Solve each trig equation on the interval 0 <or= to x < 2pi.

(1) 4 cos^2 (x) - 3 = 0

(2) tan(x/2) = sqrt{3}

What exactly do they mean by solving the equation
"on the interval 0 <or= to x < 2pi"?

2. Originally Posted by magentarita
Solve each trig equation on the interval 0 <or= to x < 2pi.

(1) 4 cos^2 (x) - 3 = 0

(2) tan(x/2) = sqrt{3}

What exactly do they mean by solving the equation
"on the interval 0 <or= to x < 2pi"?
Hello magentarita,

Since trig functions are cyclic i.e $f(x)=f(x+h)$(Ex: $\sin(x)=\sin(x+2\pi)$) for some fixed value of h the above equations will have infinite solutions. When given an interval it reduces the number of solutions.

The first one can be written as the difference of squares and we can use the zero product property

$4\cos(x)-3=0 \iff [2\cos(x)]^2-(\sqrt{3})^2=0 \iff [2\cos(x)-\sqrt{3}][2\cos(x)+\sqrt(3)=0$

setting each factor above equal to zero we can solve the equation

$2\cos(x)-\sqrt{3}=0 \iff \cos(x)=\frac{\sqrt{3}}{2} \iff x=\frac{\pi}{6} \mbox{ or } x=\frac{11\pi}{6}$

And the other factor

$2\cos(x)+\sqrt{3}=0 \iff \cos(x)=-\frac{\sqrt{3}}{2} \iff x=\frac{5\pi}{6} \mbox{ or } x=\frac{7\pi}{6}$

Good luck

3. Hey...

Why did you include a square root of 3 in question one when there is no such number there? The square root of 3 applies to question 2 only.

4. Originally Posted by magentarita
Why did you include a square root of 3 in question one when there is no such number there? The square root of 3 applies to question 2 only.

3 is not a perfect square. However...

$3=(\sqrt{3})^2$

Now the above equation can be written as the difference of squares and we can use the formula

$a^2-b^2=(a-b)(a+b)$

I hope this clears it up.

Good luck.

TES

5. 0 <or= to x < 2pi.

this simply means the value of x must lie within that aforementioned range

6. Thank you...

I thank all of you for your help.