Re: More trigonometric equations!!!

On Thu, 29 Jan 2009 19:07:32 -0800 (PST), Albert
<albert.xtheunknown0@xxxxxxxxx> wrote:

Thanks to everyone so far! You're really helping me learn this stuff
on my own + with a little guidance.

:)

Example 8. Solve the following equation for 0 degrees <= x <= 360
degrees:
cos^2(x) + 5sin(x)cos(x) = 3

I've tried dividing by A square of a trig function. I've also tried
substituting 3 = 3sin^2(x) + 3cos^2(x). But what I've found is that
the term involving sin(x)cos(x) is always present and prevents any
quadratic equation from appearing. What should I next try to do?

While there are various ways to solve your equation, you might want to
"expand your horizons" a bit and consider the following.

If we let u = tan(x) and express sin(x) and cos(x) in terms of u,
we find that

sin(x) = u/sqrt(1+u^2) and cos(x) = 1/sqrt(1 + u^2)

These relations are very easy to derive with a little thought.


With these substitutions, we have for your problem

1/(1 + u^2) + 5 u/(1 + u^2) - 3 = 0

Simplifying,

1 + 5u - 3 (1 + u^2) = 0

3 u^2 - 5u + 2 = 0

Solving the quadratic,

u = 1 and u = 2/3


Substituting tan(x) for u,

tan(x) = 1 and tan(x) = 2/3

and so the two solutions in quadrant I are

x = atan(1) = pi/4 and x = atan(2/3)

The other two solutions in quadrant III are

x = pi + pi/4 = 5 pi/4 and x = pi + atan(2/3)


Expressed in degrees, the solution set is
x ~= 33.69007
x = 45
x ~= 213.69007
x = 225

.