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Can someone show me the steps to solving this? Thank you!

8 cos^2 (theta) - 3 = 1

2007-10-17 11:42:15 · 11 answers · asked by Anonymous in Science & Mathematics Mathematics

Thank you all, you are very helpful!

...I wish I could figure these out more easily. Suffice to say math is NOT my subject, Trig even more so!

2007-10-17 11:55:07 · update #1

11 answers

8 cos ² θ = 4
cos ² θ = 1/2
cos θ = ± 1 / √2
θ = π/4 , 3π/4 , 5π/4 , 7π/4

2007-10-17 21:19:20 · answer #1 · answered by Como 7 · 0 0

8cos^2(X)-3=1
8cos^2(X)=4
cos^2(X)=4/8
(cos^2(x))^1/2=(1/2)^1/2
cos(x)=(1/2)^1/2
x=cos((1/2)^1/2)^-1)
x=45 degrees

* I set x equal to theta
* anything raised to 1/2 power is the same as square root
* x is equal to the cosine inverse of the square root of 1/2

2007-10-17 12:08:31 · answer #2 · answered by george p 2 · 0 0

8 cos^2 (theta) - 3 = 1
=> 8 cos^2 (theta) = 4
=> cos^2 (theta) = 1/2
=> cos(theta) = +/- 1/sqrt2
=> theta = 45,135, 225, 315 deg

2007-10-17 11:46:25 · answer #3 · answered by harry m 6 · 1 0

8COS^2(THETA) = 1+3 = 4
cos^(theta) = 4/8 = 1/2
cos(theta) =+- sqrt(1/2)
theta = cos inverse(+-sqrt(1/2))
theta = +-45 degrees

2007-10-17 11:47:58 · answer #4 · answered by whitney_llove 3 · 0 0

if ur allowed a calculator, it's fairly easy. whenever you want to solve for something you just have to move everything else out of the way, keeping in mind to do the same thing to both sides of the equation.around until you've got it. for instance i would move the 3 over by adding it, then divide both sides by 8 to get rid of that term, then take the square root of booth sides, then take the inverse of that number. the reason i did it in that order is because you usually save the "hardest" things to get rid of last, so adding subtracting is easiest, followed by multiplication and division, followed by powers, followed by cosines and sines.

2007-10-17 11:49:29 · answer #5 · answered by keyahnoo 2 · 0 0

Add 3 to both sides. Divide both sides by 8. This gets cos^2(θ) on one side. Then take the square root (don't forget the ±) and arccosine.

2007-10-17 11:46:17 · answer #6 · answered by Anonymous · 0 0

Ok this is simple. First let x=cos^2(theta)

8x-3=1

So x=0.5

cos(theta) = 0.25 (plus or minus)

Just take the inverse cosine of 0.25 and you get

k + n*pi

2007-10-17 11:50:23 · answer #7 · answered by Anonymous · 0 0

you need to isolate Theta
8cos^2(Theta) = 1 + 3
cos^2(theta) = 4 / 8
cos(theta) = sqrt(1/2)
thetha = arccos (sqrt(1/2)) = pi / 4 = 45°

2007-10-17 11:48:56 · answer #8 · answered by landonastar 3 · 0 0

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2016-10-04 01:17:00 · answer #9 · answered by ? 4 · 0 0

8cos^2(th) = 4

cos^2(th) = 0.5

cos(th) = 1/sqrt(2)

th=pi over 4 radians (or 45 degrees)

2007-10-17 11:45:35 · answer #10 · answered by Anonymous · 1 0

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