trigonometry?

Question

Hello! If there is anyone here smart enough to solve this equation, I'll be very glad.

3cos2x - 5cosx = 1

helpppp!!! Pleeeaaase!!!

Thank you.

Answer 1

3cos(2x) - 5cos(x) = 1

I assume you actually mean a "2" and not a squared. With that said, remember the double angle formula:

cos(2y) = cos^2(y) - sin^2(y) = 2cos^2(y) - 1

Use this on cos(2x), and that would give us

3[2cos^2(x) - 1] - 5cos(x) = 1
6cos^2(x) - 3 - 5cos(x) = 1

Move everything to the left hand side, and sort in descending powers of cosine.

6cos^2(x) - 5cos(x) - 4 = 0

Now, we factor this like a quadratic, because that's what this is; if you don't believe me, I'll let z = cos(x). Then we have

6z^2 - 5z - 4 = 0

(3z - 4) (2z + 1) = 0

Which means

3z - 4 = 0, 2z + 1 = 0
z = {4/3, -1/2}

BUT we let z = cos(x), so what we're really solving for is

cos(x) = 4/3
cos(x) = -1/2

I'm going to assume that 0 <= x < 2pi

We can solve cos(x) = -1/2; x = {2pi/3, 4pi/3}

cos(x) = 4/3 is actually not possible. Recall that the range of cos(x) is [-1, 1], and 4/3 exceeds 1. Therefore, our only solutions are

x = {2pi/3, 4pi/3}

If we wanted the general unrestricted solutions, it would be

x =
{2pi/3 + 2k(pi) | k an integer} U
{4pi/3 + 2k(pi) | k an integer}

Answer 2

Use the double angle formula cos(2x) = 2cos²x - 1

3cos2x - 5cosx = 1
3(2cos²x - 1) - 5cosx - 1 = 0
6cos²x - 3 - 5cosx - 1 = 0
6cos²x - 5cosx - 4 = 0
(3cosx - 4)(2cosx + 1) = 0
cosx = 4/3, -1/2
4/3 is eliminated as impossible since cos x cannot be greater than one.

cos x = -1/2
x = {2π/3 + 2πn, 4π/3 + 2πn} where n is an integer.

Answer 3

use the trig identity: cos(2x) = 1 - 2 sin^2(x)

So 3*(2cos^2x-1) - 5cosx= 1
Turn this into a polynomial

6cos^2x - 5cosx - 3 - 1 = 0
6 cos^2x - 5cosx - 4 = 0
and solve for cosx using the quadratic formula!

Answer 4

Remember the identity "cos 2x = 2(cos x)^2 -1".
The equation boils down to:

6 (cos x)^2 - 5 cos x - 4 = 0

this is a simple quadratic, hope you can take it the rest of the way!

Answer 5

Yes.

You use the double angle identity cos(2x) = 2cos(x)^2-1 then solve this quadratic for cos(x) and then use special angles. Maple answer:

> solve(3*cos(2*x)-5*cos(x)=1,x);

2/3 Pi, arccos(4/3)

Note: arccos(4/3) is not a real number, so we (usually, depending on the class) ignore this one.

Answer 6

3cos2x-5cosx=1
3(2cos^2x-1)-5cosx=1
6cos^2x-5cosx-3-1=0
6cos^2x-5cosx-4=0
6cos^2x+3cosx-8cosx-4=0
(3cosx-4)(2cosx+1)=0
cosx=4/3 or -1/2
cosx=4/3 is not admissible as cosx max=1
cosx=-1/2
cosx=cos2pi/3
x=2npi++/-2pi/3

Answer 7

i cannot. i made straight a's when i took that back in 1963. had it down cold. that is a simple equation to solve but i am sorry i tried and cannot remember. that is what happens as time goes by. i cannot remember this but a kiss is still a kiss. trig evaporates.

Answer 8

i'm not gonna answer it, you are.
Remember the Pathagarian Identity.
(sinx)^2+(cosx)^2 =1

Answer 9

cosx = -0.5
which means that x = 3*pi/4