Question

solve 4sin^2 x=4cosx+1

Answer 1

No plan? This one is NOT an identity. This one is conditional.

Could you solve it if it were this? \(4x^{2} = 4x + 1\) -- No trig, here, just algebra.

Answer 2

sorry, all solutions are within [0,2pi)

Answer 3

You didn't answer my question. The restriction you suggest is common.

Answer 4

i don't quite understand the question you're asking

Answer 5

Okay, let's see if I can explain it better...

You are walking down the street, minding your own business. An alien life form appears before you. The life form says, via its universal translator, "I will give you an algebra problem. If you cannot solve it, we will destroy your planet."

Solve for x, 4x^2 = 4x + 1

Well, will we survive, or not?

Answer 6

probably not.

Answer 7

Seriously? It's a Quadratic Formula at worst.

4x^2 - 4x - 1 = 0

x = [4 + sqrt(16 - 4(4)(-1))]/(2*4) or [4 - sqrt(16 - 4(4)(-1))]/(2*4)

Agreed?

Answer 8

got you. i was approaching the problem completely different so i was confused why my answer didnt make any sense

Answer 9

how do i relate this back to my original problem ?

Answer 10

No worries. It is often a bit of a shock to get an algebra problem in the middle of a trigonometry problem. This is what we have.

4sin^2 x=4cosx+1

Use the Pythagorean identity to transform to cosine only.

4(1 - cos^2(x))=4cosx+1

So far, so good?

Answer 11

yep

Answer 12

Now, rearrange so that it looks like a normal quadratic equation.

4(1 - cos^2(x)) = 4cosx + 1

4 - 4cos^2(x) = 4cosx + 1

4cos^2(x) + 4cosx - 3 = 0

Now, it's not QUITE a normal Quadratic Equation, but it is so close that if we write

cos(x) =

instead of

x =

That is the ONLY difference.

Can you solve it for cos(x)? It is a Quadratic Formula, at worst. Perhaps it can be factored.

Answer 13

for this part 4 - 4cos^2(x) = 4cosx + 1, where did the 4 - come from ?

Answer 14

sorry just realized you multiplied the 1 inside haha

Answer 15

Good. Do you see the final structure?

Answer 16

-4+-sqrt4^2-4(4)(-3)/2*4

Answer 17

Really... Did you try to factor it?

4cos^2(x) + 4cosx - 3 = 0

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

Answer 18

how do you use the quadratic formula after factoring it ?

Answer 19

You don't. The quadratic formula is for stubborn things that cannot be factored.

You now use the zero property of multiplication.

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

2cos(x) + 3 = 0
cos(x) = -3/2 <== Well, that's just silly. Discard it. Do you see why?

2cos(x) - 1 = 0
cos(x) = 1/2 ==> x = pi/3, 5pi/3

Answer 20

that was a lot easier than suspected. thank you once again :)

Answer 21

Once you see that you need to remember your algebra, you are on your way with this one!

If you see those aliens, don't let them intimidate you!

Answer 22

haha you got it