I expanded it as much as I could:
2cos^8x - 8cos^6x + 12cos^4x - 7cos^2x + 6cosx +5 = 0

But IF you assume its sin^2x, then the equation can be solved nicely,
giving (cosx + 1 )(cosx - 7) = 0

Choose your destiny.

Reply 4

ImABigOldTurd

OP

5

Original post by peterith

I expanded it as much as I could:
2cos^8x - 8cos^6x + 12cos^4x - 7cos^2x + 6cosx +5 = 0

But IF you assume its sin^2x, then the equation can be solved nicely,
giving (cosx + 1 )(cosx - 7) = 0

Choose your destiny.

hahahahaha alright thanks I think I'll just give up on it

Reply 5

tiny hobbit

15

Original post by ImABigOldTurd

for 0 less than or equal to x which is less than 2pi

(3+cosx)²=4-2sin⁸x

what can I do about that horrible power?

Consider the largest and smallest possible values of the left hand side and of the right hand side as x varies.

You should find that there is only one possible valus of each side of the equation.

It comes from an Oxford MAT paper doesn't it?

(edited 9 years ago)

Reply 6

WhiteGroupMaths

9

This probably needs to be solved by inspection rather than brute force expansion.

Both the LHS and RHS share the same periodicity, and in fact the equation can only hold if both sides yield an integer value.

The broad solution set would actually be (2k+1) pi , where k is any chosen integer value. You can certainly decipher the answer within the assigned range as given in the question, and as rightly mentioned by tiny hobbit, there is only one such answer.

Hope this helps. Peace.

Reply 7

james22

16

Do what tiny hobbit says, this cannot be solved by standard methods (easily).

Reply 8

ztibor

10

Original post by ImABigOldTurd

for 0 less than or equal to x which is less than 2pi

(3+cosx)²=4-2sin⁸x

what can I do about that horrible power?

The range of LHS as function is [4,16]
The range of RHS as function is [2,4]

THe equation is true only when the value is 4 in both side (there is only one common value)

(3+cosx)^2=4 -> cosx=-1 -> x= 180 + m*360

4-2sin^8x =4 ->sinx=0 -> x=n*180

So there is one solution between 0 and 2pi

Reply 9

ImABigOldTurd

OP

5

Original post by tiny hobbit

Consider the largest and smallest possible values of the left hand side and of the right hand side as x varies.

You should find that there is only one possible valus of each side of the equation.

It comes from an Oxford MAT paper doesn't it?

it was in a cruel mixture of different questions but yeah some are from STEP