Mathematicians... I need help!

I'm assuming you are trying to solve for x over the interval [0, 2pi].

Let's look at the first problem:

cosx - cos2x = 0.

Use the Double-Angle Formula to replace cos2x with 2cos^2x -1. So we now have:

cosx - (2cos^2x -1) = 0.

Distribute and rearrange to get -2cos^2x + cosx +1 = 0

This factors out to become (-2cosx - 1)(cosx - 1) = 0

Set each product equal to zero and solve for cosx. We now get cosx = -1/2 and 1.

Should be easy enough to find the values of x. When cosx = -1/2 then x = 5/6pi or 7/6pi. When cosx = 1 then x = 0 or 2pi.

The second problem has a little more algebra involved but it works the same way.

tan2x - 2cosx = 0

Substitute tan2x using the Double-Angle Formula. We now have:

[2tanx/(1-tan^2x)] - 2cosx = 0

Factor out a 2 and we get 2{[tanx/(1-tan^2x)] - cosx} = 0

We know from basic algebra that if a times b equals 0 then either a = 0 or b = 0. From above, since 2 does not equal 0 we can drop it and concentrate on the stuff inside the braces. We now have:

[tanx/(1 - tan^2x)] - cosx = 0

Next multiply both sides by (1 - tan^2x) to get:

tanx - cosx(1 - tan^2x) = 0

Distribute: tanx - cosx + tan^2xcosx = 0

Subsitute tan^2x with sin^2x/cos^2x

tanx - cosx + (sin^2x/cos^2x)cosx = 0

This simplifies to:

tanx - cosx + sin^2x/cosx = 0

Multiply both sides by cosx. (note: tanx times cosx equals sinx).

sinx - cos^2x + sin^2x = 0

Now we use the Double-Angle Formula to get:

sinx - 1 + 2sin^2x = 0

Rearrange to:

2sin^2x + sinx - 1 = 0

Use the quadratic formula to solve for sinx. Once that is done we should have sinx = 1/2 or -1.

Solving for x gives us the solutions pi/6, 5/6pi or 3/2pi.

If you need x in degrees instead of radians then just multiply the solutions by 180/pi.

 

Using the tangent double-angle formula throws out the perfectly good solution x = pi/2.. to be a little more careful, try

tan2x - 2cosx = 0
=> (sin2x - 2cosxcos2x)/cos2x = 0
=> 2cosx(2sinx - 1)(sinx + 1)/cos2x = 0

which gives all four solutions in [0,2pi] (after checking each for asymptotic behaviour in the denominator). Hope this helps

 

I haven't done any of this shizz since high school calculus and I signed up for two math classes this semester. Yay for not remembering any of this....

 

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