# Trig equations

• Apr 5th 2010, 07:39 AM
darksupernova
Trig equations
Hello,

I am having a bit of trouble solving this eqn:

tan (x) + cot (x) = 2sec (x)
For all angles between 0 and 360 degrees.

Thank you!

Max
• Apr 5th 2010, 08:33 AM
Sudharaka
Quote:

Originally Posted by darksupernova
Hello,

I am having a bit of trouble solving this eqn:

tan (x) + cot (x) = 2sec (x)
For all angles between 0 and 360 degrees.

Thank you!

Max

Dear darksupernova,

$\displaystyle tanx+cotx=2secx$

$\displaystyle \frac{1}{sin(x)cos(x)}=2secx$

$\displaystyle 2sinx=1$

$\displaystyle sinx=\frac{1}{2}$

Since x is inbetween 0 and 360 degrees,

$\displaystyle x=30^0$

• Apr 5th 2010, 09:27 AM
darksupernova
Ah very clever!

Thank you very much!

Im now a but stuck on this:

tan (x) + 3cot (x) = 5sec (x)

Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work?
• Apr 5th 2010, 09:38 AM
Quote:

Originally Posted by darksupernova
Hello,

I am having a bit of trouble solving this eqn:

tan (x) + cot (x) = 2sec (x)
For all angles between 0 and 360 degrees.

Thank you!

Max

$\displaystyle tanx+cotx=\frac{sinx}{cosx}+\frac{cosx}{sinx}=\fra c{sinx}{sinx}\ \frac{sinx}{cosx}+\frac{cosx}{cosx}\ \frac{cosx}{sinx}$

$\displaystyle =\frac{sin^2x+cos^2x}{sinxcosx}$

$\displaystyle \Rightarrow\ \frac{1}{sinxcosx}=\frac{2}{cosx}\ \Rightarrow\ \frac{1}{sinx}=2\ \Rightarrow\ sinx=\frac{1}{2}$

The vertical co-ordinate of the angle is 0.5 in the unit circle, centre (0,0).
This corresponds to two angles,

$\displaystyle x=sin^{-1}\left(\frac{1}{2}\right),\ x={\pi}-sin^{-1}\left(\frac{1}{2}\right)$

$\displaystyle x=30^o,\ x=150^o$
• Apr 5th 2010, 09:49 AM
Quote:

Originally Posted by darksupernova
Ah very clever!

Thank you very much!

Im now a but stuck on this:

tan (x) + 3cot (x) = 5sec (x)

Very similar but the cos^2 {x) + sin^2 (X) = 1 trick doesnt work?

$\displaystyle \frac{sinx}{cosx}+\frac{3cosx}{sinx}=\frac{5}{cosx }$

$\displaystyle \frac{sin^2x+3cos^2x}{cosxsinx}=\frac{5}{cosx}$

$\displaystyle \frac{sin^2x+cos^2x+2cos^2x}{cosxsinx}=\frac{1+2co s^2x}{cosxsinx}=\frac{5}{cosx}$

$\displaystyle \frac{1+2cos^2x}{sinx}=5$

$\displaystyle 2cos^2x+1=5sinx$

$\displaystyle 2-2sin^2x-5sinx+1=0$

$\displaystyle 2sin^2x+5sinx-1=0$

$\displaystyle sinx=\frac{-5\pm\sqrt{25+8}}{4}$

x is the inverse sine of the 2 results.
• Apr 5th 2010, 10:02 AM
darksupernova
Another clever little trick on line three, hadnt thought of that, thankyou very much! Now that ive seen these tricks i wont forget them!