Please help me to solve x , 0 <= x <= 360, where

3 tan x - 2 + 3 sec x = 2 cosec x

Thank you

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- Jun 20th 2007, 12:38 AMNicoSolve x (tan sec and cosec)
Please help me to solve x , 0 <= x <= 360, where

3 tan x - 2 + 3 sec x = 2 cosec x

Thank you - Jun 20th 2007, 05:15 AMtopsquark

<-- Clear the fractions

We want to get this equation in terms of a single trig function. We can do that by employing , so we need to do the following:

<-- Square both sides

Now . (If we hadn't squared the equation we'd've had to choose which quadrant x is in.)

Now let

This is ugly. BUT, if we use the rational root theorem we know that all the possible rational roots are at . It turns out that is a rational zero. So by division we get that

We may apply the rational root theorem again to find that is again a zero of the cubic polynomial, or you may note that we may factor the cubic polynomial by grouping:

So:

So we have that

and

The solution generates

The other two are not so nice. We obtain:

which has only the approximate solutions:

and

which has only the approximate solutions:

-Dan

PS Whoops! I almost forgot. We need to check each of these solutions in the original equation to be sure that they ARE actually solutions. I get that only the solutions work.

Graphically solving the problem makes it appear that is also a solution. However this x value can't be a solution because is undefined.

-Dan - Jun 20th 2007, 05:29 AMNico
Thank you, what about:

3 sin x / cos x - 2 + 3 /cos x = 2 / sin x

2 sin x.sin x - 2 sin x cos x = 3 sin x = 2 cos x

3 sinx (sin x +1) = 2 cos x (sin x +1)

3 sinx = 2 cos x

tan x = 2/3

x = 33,69 and 213,69 - Jun 20th 2007, 05:32 AMtopsquark