Trig equation:?

Question

Find the values where 2 + cos2x = 3cosx in the interval [0, 2π] (i.e. including 0 and 2π) ... i tried about a million different things and now i've got a head ache.... please help ^^

p.s. i know the answer is 0, π/3, 5π/3, and 2π, but i did that graphically and i need to do it algebraically.

Answer 1

you need to convert it to one trigonometric function of the same angle (or argument)...

in this case, it is easier to convert cos(2x) into cos(x)...
note: cos(2x) = 2cos^2(x) -1

thus
2 + 2cos^2(x) -1 = 3 cos(x)
2cos^2(x) - 3cos(x) + 1 = 0 ... factor this
(2cosx -1) (cos x -1) = 0

cosx = 1/2 ... x = π/3, 5π/3

cosx = 1 ..... x = 0, 2π
these are the solutions...



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Answer 2

2 + cos2x = 3cosx
2 + 2(cosx)^2 - 1 = 3cosx ==> double angle identity of cosine
2(cosx)^2 - 3cosx + 1 = 0 ==> quadratic form in cosx variable
(2cosx - 1)(cosx - 1) = 0

Then we get :
cos x = 1/2
x = π/3 and 5π/3

OR

cosx = 1
x = 0 and 2π

Answer 3

2 + 2 cos ² x - 1 = 3 cos x
2 cos ² x - 3 cos x + 1 = 0
(2 cos x - 1)(cos x - 1) = 0
cos x = 1 / 2 , cos x = 1
x = π/3 , 5π/3 , 0 , 2π

Answer 4

Note that

cos2x = 2(cosx)^2-1

So the equation reduces to

2(cosx)^2-3cosx+1 = 0, or

(2cosx-1)(cosx-1)=0, or

cosx=1/2, 1

So there are several feasible values for x, and they are 0, pi/3, 5pi/3 and 2pi.

Answer 5

2 + cos(2x) = 3cos(x)

Use the double angle identity for cosine.

cos(2x) = 2cos^2(x) - 1

2 + 2cos^2(x) - 1 = 3cos(x)

Moving everything to the left hand side, we get

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

Simplifying, we get

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

Which we factor as a quadratic.

[2cos(x) - 1] [cos(x) - 1] = 0

Equate each factor to 0 and solve.

2cos(x) - 1 = 0
cos(x) - 1 = 0

2cos(x) = 1
cos(x) = 1

cos(x) = 1/2
cos(x) = 1

Our interval is [0, 2&pi], so our answers are

1) cos(x) = 1/2
x = {&pi/3, 5&pi/3}

2) cos(x) = 1
x = {0, 2&pi}

So our solutions, like you said, are

x = {0, &pi/3, 5&pi/3, 2&pi}

Answer 6

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