Question

solve tan3x - sec3x = 5

Answer 1

rearange so you have \[\ \tan(3x) -5 = \sec (3x)\]\[\ ( \tan(3x) -5)^2 = (\sec(3x))^2\]\[\ \tan^2 (3x) -10\tan(3x) - 25 =\sec^2 (3x)\] How far I've gotten.

Answer 2

Im stuck here :|

Answer 3

\(\Large sec^23x-tan^23x=1\)

Answer 4

How did you get that? :o

Answer 5

\(\Large sec^2A-tan^2A=1\)

famous identity

http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Answer 6

Oh this, yes.

Answer 7

I still don't understand how to apply it :(

Answer 8

Try factoring a \(\sec x\).

Answer 9

\[\begin{align*}
\tan3x-\sec3x&=5\\
\frac{\sin3x}{\cos3x}-\frac{1}{\cos3x}&=5\\
\frac{1}{\cos3x}\left(\sin3x-1\right)&=5\\
\sin3x-1&=5\cos3x
\end{align*}\]
might be easier to work with.

Answer 10

Ohh I see I see.

Answer 11

\(\large\tt \begin{align} \color{black}{ -10\tan(3x) + 25 =\sec^2 (3x)-\tan^2 (3x) \\
-10\tan(3x) +25 =1\\
tan(3x)=\dfrac{12}{5}\\
x=\dfrac{1}{3}arctan\dfrac{12}{5}\\
\approx 0.39}\end{align}\)

Answer 12

Oh is that what you were referring to -_- I goofed.

Answer 13

@Jhannybean there should be\(+25\) in ur equation

Answer 14

Yes you are right :) Haha.

Answer 15

\[\rm \tan(3x)=\dfrac{12}{5} \implies 3x = \arctan\left(\frac{12}{5}\right) + n\pi \implies x = \frac{1}{3}\left[ \arctan\left(\frac{12}{5}\right) + n\pi\right]\]

Answer 16

why not \(\Large 2n\pi\)

Answer 17

period of tan is just \(\pi\), so it repeats after every \(\pi\) :
\[\tan(x) = \tan(x+\pi) = \tan(x+\pi+\pi) = \tan(x+\pi+\pi+\pi+\cdots)\]