Question

solve sec theta-cosec theta=4/3

Answer 1

\[\left\{\theta =-\text{ArcSec}\left[\frac{2}{3} \left(1-\sqrt{7}\right)\right],\theta =\text{ArcSec}\left[\frac{2}{3} \left(1+\sqrt{7}\right)\right]\right\} \]

Answer 2

???

Answer 3

My goodness, your teachers torture you.

Answer 4

Just to make it easier, I'm going to rewrite the problem in x

I.e., sec x-cosec x=4/3

Let u = cosec x. Then 1/u = sin x and hence cos x = +- sqrt(1-1/u^2). Thus

1/cos x = sec x = 4/3 + u

Squaring

\[\frac{1}{1 - 1/u^2} = \frac{u^2}{u^2 - 1} = (u + 4/3)^2\]

Answer 5

Thus
\[u^2 = (u^2 - 1)(u+4/3)^2 = (u^2 - 1)(u^2 + 8/3u + 16/9) = ...\]

Answer 6

So you will end up with a fourth order equation in u and you have to look for roots.

Answer 7

Like I say, your high school teachers are a bit sadistic.

Answer 8

oh is ther any othr way i cannot use fourth order equations

Answer 9

http://www.wolframalpha.com/input/?i=solve+sec+theta+-+cosec+theta+%3D+4%2F3+

I used by personal Home Edition of Mathematica 8 for the two solutions I posted above as a first responder. There were actually another two solutions with complex zeros.

Answer 10

Yes ... take the equation I wrote down above and subtract u^2 from both sides and you get

1/9 ( u^4 + 8u^3 - 2u^2 - 24u - 16)

Factorize that and you have

1/9 (3u^2 + 4u + 2)(3u^2 + 4u - 8)

So this is equal to zero means that either

3u^2 + 4u + 2 = 0 OR 3u^2 + 4u - 8 = 0

Now, in standard notation, the discriminent of the first equation
b^2 - 4ac = 16 - 24 = -8 < 0 and therefore the equation has no real solutions.

The other equation does have two roots:

\[u = \frac{2}{3} ( -1 \pm \sqrt{7})\]

Thus
\[\theta = arcsec \left( \frac{2}{3}(-1 \pm \sqrt{7}) \right)\]

Answer 11

So this is how you can reduce the 4th order equation to quadratics, which you know how to handle.

Answer 12

Sigh.
Correction: the equation is:
1/9 ( 9u^4 + 8u^3 - 2u^2 - 24u - 16)