exact solutions to trig equation Cos^2(x- pi/3) = 1/2

**bakerkhojah** · August 9th 2012, 07:30 PM

hello everyone im new to this forum sorry didnt have time to introduce myself as my exams are coming pretty soon and i wish someone could help me with this problem about exact solutions to trig equations...

so im supposed to use this trig equation to find exact solutions without using a GDC its an example in the book but im not sure how to arrive at the solution they have so here is an image of the example in the book and the solution....

okay so i followed till they got that + or -sqrt of 2/2 is pi/4 then i dont understand how they arrived at the equation x-(pi/3) = pi/4 + k x pi/2. based on what is it pi/2? please explain im lost.

**Prove It** · August 9th 2012, 07:43 PM

Originally Posted by bakerkhojah

hello everyone im new to this forum sorry didnt have time to introduce myself as my exams are coming pretty soon and i wish someone could help me with this problem about exact solutions to trig equations...

so im supposed to use this trig equation to find exact solutions without using a GDC its an example in the book but im not sure how to arrive at the solution they have so here is an image of the example in the book and the solution....

okay so i followed till they got that + or -sqrt of 2/2 is pi/4 then i dont understand how they arrived at the equation x-(pi/3) = pi/4 + k x pi/2. based on what is it pi/2? please explain im lost.

We want all the angles in the unit circle that give the cosine value (the red line segments) as $\displaystyle \begin{align*} \pm \frac{1}{\sqrt{2}} \end{align*}$ . You should know that the first value is $\displaystyle \begin{align*} \frac{\pi}{4} \end{align*}$ , and can you see that all the other angles are separated by an angle of $\displaystyle \begin{align*} \frac{\pi}{2} \end{align*}$ ? Once you have those four values, then you keep adding the integer multiples of $\displaystyle \begin{align*} 2\pi \end{align*}$ because you could keep going around the circle.

**bakerkhojah** · August 10th 2012, 09:51 AM

thank you for your help i got the first 3 solutions right according to your way but i kind of cant find the way to do the 4rth as it corresponds to 7pi/4 and when u add pi/3 to it you get 25pi/12 which is not in the interval 2pi since it has a domain of 0 </= x </= 2pi and the forth solution is basically pi/12, im sorry im asking so many questions i just want to know how to do it haha ur way helped me in the first three solutions but i cant figure out how to do the forth...

**Prove It** · August 10th 2012, 06:01 PM

Originally Posted by bakerkhojah

thank you for your help i got the first 3 solutions right according to your way but i kind of cant find the way to do the 4rth as it corresponds to 7pi/4 and when u add pi/3 to it you get 25pi/12 which is not in the interval 2pi since it has a domain of 0 </= x </= 2pi and the forth solution is basically pi/12, im sorry im asking so many questions i just want to know how to do it haha ur way helped me in the first three solutions but i cant figure out how to do the forth...

When you add all the integer multiples of $\displaystyle \begin{align*} 2\pi \end{align*}$ you have to add all the NEGATIVE integer multiples as well (which is equivalent to going around the circle in the other direction).

**bakerkhojah** · August 13th 2012, 01:29 PM

ahh thank you makes so much sense now!

i have another problem and if u could help me with it plz,
so i want to find the exact solutions to the problem,

sin2x=sprt3/2 for the interval 0</= x </=2pi

and by finding the inverse sine i find it being pi/3 and by solving for x i get also pi/6 those are the two first exact solutions now since its a positive square root thus we are looking for where sine is positive which means that the other solutions are 2pi/3 and 5pi/6 however according to the book that is wrong and the answer is pi/3, pi/6, 7pi/4, and 4pi/3, how could this be?!

**skeeter** · August 13th 2012, 02:33 PM

in future, one question per thread ... start a new one.

$0 \le x \le 2\pi$

$0 \le 2x \le 4\pi$

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

$2x = \frac{\pi}{3} \, , \, \frac{2\pi}{3} \, , \, \frac{7\pi}{3} \, , \, \frac{8\pi}{3}$

$x = \frac{\pi}{6} \, , \, \frac{\pi}{3} \, , \, \frac{7\pi}{6} \, , \, \frac{4\pi}{3}$

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