Asked in Math and Arithmetic, Algebra, Calculus
Math and Arithmetic
Algebra
Calculus

How do you solve the equation for x 2cos x squared plus 3sin x equals 0? 04/28/2010

Well, let's see.

2 cos2(x) + 3 sin(x) = 0

We're going to use the only powerful magic trick we know right now, right at the beginning.

Then, after that, we'll have no tricks left, and it'll just be a slog from there to the end.

The Trick: [ cos2(x) ] is the same thing as [ 1 - sin2(x) ]. That's the only trig trick we have.

Substitute that in the original equation where you see cos2(x) :

2 [ 1 - sin2(x) ] + 3 sin(x) = 0

Now, since we have a lot of slogging to do, let's make it a little easier on ourselves.

Wherever we have sin(x), let's write 'S' instead. At the end, when we figure out what 'S' is,

we'll go back to calling it "sin(x)" again .

2 (1 - S2) + 3 S = 0 . . . (See ? Isn't that a lot easier to write ?)

Eliminate the parentheses on the left side:

2 - 2 S2 + 3 S = 0

Re-arrange it a little:

-2 S2 + 3 S + 2 = 0

Just for convenience, multiply each side by -1:

2 S2 - 3 S - 2 = 0

Factor the left side:

(2S + 1) (S - 2) = 0

If either factor is zero, we get a solution for the equation:

2S + 1 = 0 . . . . . 2S = -1 . . . . . S = -1/2

and

S - 2 = 0 . . . . . S = 2

Now that we have actual numbers for 'S', we can call it "sin(x)" again,

and figure out what the angle 'x' is.

The first solution: sin(x) = -1/2 . . . . . x = 210° , 330°

Second solution: sin(x) = 2 . . . . . That's a nonsense solution. The sine of anything can't be greater than 1.

So we're done.