# solving trigonometric equation

• Sep 6th 2008, 01:05 PM
myoplex11
solving trigonometric equation
solve tanx+secx=2cosx
(sinx/cosx)+ (1/cosx)=2cosx
(sinx+1)/cosx =2cosx
multiplying both sides by cosx
sinx + 1 =2cos^2x
sinx+1 = 2(1-sin^2x)
2sin^2x + sinx-1=0
(2sinx+1)(sinx-1)=0
x=30 x=270
but if i plug 270 back into the original equation i get undefined because tan 270 is undefined where did i go wrong?
• Sep 6th 2008, 01:10 PM
Matt Westwood
Because neither of tan x nor sec x are defined at 270 degrees, the original derivation of your quadratic in sines does not apply (good job btw) so it's not a valid solution.

(Note you also have 30 + integer multiples of 360 degrees as solutions as well.)

Try plotting tanx + secx and 2 cosx on the same graph and see what it looks like. Note the singular point at $\pm 90 + 360n$ to see what I mean.
• Sep 6th 2008, 01:27 PM
myoplex11
so i do i solve it algebraically.
• Sep 6th 2008, 01:31 PM
Soroban
Hello, myoplex11!

Quote:

Solve: . $\tan x + \sec x\:=\:2\cos x$

$\frac{\sin x}{\cos x} + \frac{1}{\cos x}\:=\:2\cos x$

Note that, at this point: $\cos x \neq 0$

. . We cannot have zero in the denominator, right?

That is: . $x \:\neq\:\pm90^o,\;\pm270^o,\;\hdots$

So, should any of those values turn up later, they must be discarded.

• Sep 6th 2008, 01:31 PM
Matt Westwood
Beg pardon, was that a question?

Yes, I would solve it algebraically like you did (it works, it's fine, as long as cos theta not zero) and then I'd probably plot some points and draw a sketch to make sure I didn't miss any points where tanx+secx intersect 2cosx.