How would you solve 4cos^2x -4cosx=2. I tried the quadratic but dont think i did it properly. answers are 111.5 degrees and 248.5 degrees

Question

kva1992 · Answer

i first tried to factor a two after i moved the 2 over so i ended up with 2(2cos^2-2cosx-1)=0

campbell_st · Answer

this is an equation that can be solved as a quadratic

divide every term by 2 and you get

$$2\cos^2(x) -2\cos(x) -1 = 0$$

now solve using the general quadratic formula

a = 2, b = 2 and c = -1

hope it helps

kva1992 · Answer

hmmm i think i did that but ill do it again

kva1992 · Answer

i keep getting stuck at (cosx= 2 +or - 2 squareroot 3) / 4

campbell_st · Answer

well you get 

$$\frac{2 \pm \sqrt{4 - 4 \times 2 \times -1}}{2 \times 2} = \frac{2 \pm \sqrt{12}}{4}$$

which simplifies to 

$$\cos(x) = \frac{1 \pm \sqrt{3}}{2}$$

kva1992 · Answer

yup i got that

kva1992 · Answer

how how do i find the degrees?

kva1992 · Answer

inverse cos?

campbell_st · Answer

just go 

$$x = \cos^{-1}(\frac{1 + \sqrt{3}}{2})$$

then the other one with a negative infront of the square root

kva1992 · Answer

oh okie cool thanks

kva1992 · Answer

really appreciate it so i was on the right track i just kept trying to find the + when that isnt possible because it exceeds the range.

kva1992 · Answer

alright so my second question would be how did they find 248.5 degrees?

campbell_st · Answer

well that's correct... there is an error with the question since to solution is larger than 1, there is no angle... 

cos(x) is always between -1 and 1

campbell_st · Answer

yep... so its only the 2nd solution 

(1 - sqrt{3})/2

so the angle is in the 2nd and 3rd quadrant...

kva1992 · Answer

but the answer sheet says its both 111.5 and 248.5

kva1992 · Answer

and i have no idea how they figured out 248.5

campbell_st · Answer

thats correct... 

$$\frac{1 - \sqrt{3}}{2} = -0.366025$$

so this is what I do... 

find the angle for the positive value 

$$x = \cos^{-1}(0.366025)$$

so the angle is 68.63 degrees

in the 2nd quadrant the angle is 

180 - 68.63

in the 3rd quadrant the angle is 

180 + 68.63 

hope this helps

kva1992 · Answer

oh yaaaa i completely forgot you arent supposed to find the negative of an inverse function duh

campbell_st · Answer

the cos ratio is negative in the 2nd and 3rd quadrants... 

2nd is always 180 - angle 

3rd is always 180 + angle

kva1992 · Answer

so if i was trying to find the sin it would be in either q1 or q4 right?

campbell_st · Answer

well all ratios are positive in the 1st quadrant

2nd only sin is positive

3rd only tan is positive

4th only cos is positive

and the angles are 

2nd  180 - angle

3rd 180 + angle 

4th 360 - angle

hope it makes sense

campbell_st · Answer

well if sin was positive its 1st and 2nd...

kva1992 · Answer

but doesnt the range influence how it works? like i remember with sine it goes from pi/2 to -pi/2

kva1992 · Answer

or am i thinking of something completely different?

ikram002p · Answer

still stuck with this ?

kva1992 · Answer

na figured the answer out just clearing up a question is all

ikram002p · Answer

^^

campbell_st · Answer

so if the range is from -pi/2 to pi/2 you are working in the 4th and 1st quadrant... 

sin is positive in the 1st and negative in the 4th

campbell_st · Answer

here is a very detailed cheat sheet on trig it may halp http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet_Reduced.pdf

kva1992 · Answer

alright cool ill look into thanks really appreciate your time and help :)