Solving Trig Equation: 6cosx + cos2x = 2sin^2(x) - 5 for 0<=x<=pi

[Vulcan]

Solving Trig Equation: 6cosx + cos2x = 2sin^2(x) - 5 for 0<=x<=pi

This one is driving me nuts, can someone put me out of my misery please?

 

[Deleted member 4993]

This one is driving me nuts, can someone put me out of my misery please?

Convert everything to sin(x) or cos(x) and continue.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[Vulcan]

Convert everything to sin(x) or cos(x) and continue.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

Am I correct so far, if so what next, if not where am i going wrong.
6cosx+cos2x+5-2sin²x =0
6cosx+cos2x+4+1-2sin²x =0
6cosx+cos2x+4+cos2x =0
6cosx+2cos2x+4=0
Divide by 2
3cosx+cos2x+2 =0
3cosx+2cos²x+1=0

Any help much appreciated

 

[stapel]

Am I correct so far, if so what next, if not where am i going wrong.

6 cos(x) + cos(2x) = 2 sin²(x) - 5

6 cos(x) + cos(2x) + 5 - 2 sin²(x) = 0

6cos(x) + cos(2x) + 4 + 1 - 2 sin²(x) = 0

6cos(x) + cos(2x) + 4 + cos(2x) = 0

6cos(x) + 2cos(2x) + 4 = 0

Divide by 2

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

3cos(x) + 2cos²(x) + 1 = 0

Any help much appreciated

Good work! Now solve the quadratic equation:

. . . . .\(\displaystyle 2\, \left[ \cos(x)\right]^2\, +\, 3\, \left[\cos(x)\right]\, +\, 1\, =\, 0\)

Applying the Quadratic Equation:

. . . . .\(\displaystyle \cos(x)\, =\, \dfrac{-(3)\, \pm\, \sqrt{\strut (3)^2\, -\, 4(2)(1)\,}}{2(2)}\)

...or simply factoring (which happens to be possible for this particular equation):

. . . . .\(\displaystyle \bigg(2\, \cos(x)\, +\, 1 \bigg)\, \bigg(\cos(x)\, +\, 1 \bigg)\, =\, 0\)

...and so forth. Then solve the two resulting trig equations.

 

[Vulcan]

Good work! Now solve the quadratic equation:

. . . . .\(\displaystyle 2\, \left[ \cos(x)\right]^2\, +\, 3\, \left[\cos(x)\right]\, +\, 1\, =\, 0\)

Applying the Quadratic Equation:

. . . . .\(\displaystyle \cos(x)\, =\, \dfrac{-(3)\, \pm\, \sqrt{\strut (3)^2\, -\, 4(2)(1)\,}}{2(2)}\)

...or simply factoring (which happens to be possible for this particular equation):

. . . . .\(\displaystyle \bigg(2\, \cos(x)\, +\, 1 \bigg)\, \bigg(\cos(x)\, +\, 1 \bigg)\, =\, 0\)

...and so forth. Then solve the two resulting trig equations.

Thanks for your help!