solve the equation sec(2x)cos(6x) + 1 = 0

[hndalama]

a. prove that cos3x = 4cos³x - 3cosx by substituting (2x + x) for 3x.

b. solve the equation sec(2x)cos(6x) + 1 = 0

I have proved the identity in part a but I need help solving part b. My attempt is as follows:

cos(3x + 3x)/cos2x + 1 = 0
cos(3x + 3x) + cos2x = 0
2cos²3x -1 + cos2x = 0
2(cos3x)(cos3x) + 2cos²x - 2 = 0
(cos3x)(cos3x) + cos²x - 1=0
(4cos³x - 3cosx)(4cos³x - 3cosx) + cos²x -1 = 0
16cos⁶x - 24cos⁴x + 10cos²x - 1 = 0
???

 

[skaa]

a. \(\displaystyle \cos(3x)=\cos(x+2x)=\cos x\cos(2x)-\sin x\sin(2x)=\cos x(2\cos^2x-1)-\sin x\cdot 2\sin x\cos x=2\cos^3x-\cos x-2\sin^2x\cos x=2\cos^3x-\cos x-2(1-\cos^2x)\cos x=4\cos^3x-3\cos x\)

 

[stapel]

a. \(\displaystyle \cos(3x)=\cos(x+2x)=\cos x\cos(2x)-\sin x\sin(2x)=\cos x(2\cos^2x-1)-\sin x\cdot 2\sin x\cos x=2\cos^3x-\cos x-2\sin^2x\cos x=2\cos^3x-\cos x-2(1-\cos^2x)\cos x=4\cos^3x-3\cos x\)

1. Please stop posting fully-worked solutions, especially without any explanation.
2. Please review carefully students' posts. For instance, in this case, the student clearly said that part (b) is at issue.

Thank you.

 

[stapel]

a. prove that cos3x = 4cos³x - 3cosx by substituting (2x + x) for 3x.

b. solve the equation sec(2x)cos(6x) + 1 = 0

I have proved the identity in part a but I need help solving part b. My attempt is as follows:

. . .cos(3x + 3x)/cos2x + 1 = 0
. . .cos(3x + 3x) + cos2x = 0
. . . 2cos²3x -1 + cos2x = 0
. . . 2(cos3x)(cos3x) + 2cos²x - 2 = 0
. . . (cos3x)(cos3x) + cos²x - 1=0
. . . (4cos³x - 3cosx)(4cos³x - 3cosx) + cos²x -1 = 0
. . . 16cos⁶x - 24cos⁴x + 10cos²x - 1 = 0
. . . ???

Ugh! This is gonna be messy, pretty much no matter how we proceed....

Obviously, they gave you the first part so you can use the second part. I agree with what you've done so far. However, when things seem only to get nastier (and the associated polynomial, 16X^6 - 24X^4 + 10X^2 - 1, does not factor), it may be time to try something else. In this case, maybe try something unexpected with the commutativity of multiplication.

. . . . .cos(6x) = cos(2(3x)) = cos(3(2x))

Now apply part (a):

. . . . .cos(3(2x)) = 4 cos^3(2x) - 3 cos(2x)

Now bring in the rest of the equation, but without doing anything with the "cos(2x)" part:

. . . . .sec(2x) cos(6x) + 1 = 0

. . . . .cos(6x) / cos(2x) + 1 = 0

Multiply through by cos(2x):

. . . . .cos(6x) + cos(2x) = 0

Then:

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

. . . . .4 cos^3(2x) - 2 cos(2x) = 0

Divide through by 2:

. . . . .2 cos^3(2x) - cos(2x) = 0

And this does factor! Now can you solve?

 

[hndalama]

. . . . .2 cos^3(2x) - cos(2x) = 0

And this does factor! Now can you solve?

that was brilliant, I solve it as follows:

cos2x(2cos²2x - 1) = 0
cos2x(sqrt(2)cos2x - 1)(sqrt(2)cos2x +1) = 0

cos2x= 1/sqrt2 and cos2x= -1/sqrt2 give the solutions x = 22.5 and 67.5 (0<x<90) which are correct

but the third factor cos2x = 0 gives a solution x= 45 which is not correct

 

[Deleted member 4993]

that was brilliant, I solve it as follows:

cos2x(2cos²2x - 1) = 0
cos2x(sqrt(2)cos2x - 1)(sqrt(2)cos2x +1) = 0

cos2x= 1/sqrt2 and cos2x= -1/sqrt2 give the solutions x = 22.5 and 67.5 (0<x<90) which are correct

but the third factor cos2x = 0 gives a solution x= 45 which is not correct

sec(2x) does not exist (hence your equation does not exist) when cos(2x) = 0

So cos(2x)=0 is excluded from the solution set.

 

[hndalama]

thank you;-)