# solving trig. equations

posted by
**Jen**
.

tan(3x) + 1 = sec(3x)

Thanks,

pretend 3x equals x

so tanx + 1 = secx

we know the law that 1 + tanx = secx

so tanx + 1 becomes secx

and... secx = secx

sec(3x) = sec(3x) [just put 3x back in for x- you don't really have to change 3x to x but it kinda makes it easier

Let 3x =u for simplicity sake

then tan u + 1 = sec u

sinu/cosu + 1 = 1/cosu

multiply by cos u

sin u + cos u = 1

square both sides

sin^2 u + 2(sinu)(cosu) + cos^2 u =1

but sin^A+cos^2Ax=1

so 1 + 2(sinu)(cos)=1

recall that sin(2A_=2sinAcosS

then sin 2u =0

or replacing the u

sin 6x = 0

by the CAST rule, 6x = 0,180º,360º

or 0, pi/2, pi in radians

so x = 0, 30º,60º or 0,pi/6,pi/3

The period of tan(3x)=180º or pi radians

so other answers can be obtained by adding mulitples of 180º or multiples of pi radians to any of the above answers.

eg. take the 60º answer, if we add 5*180 to it we get 960º

Left Side: tan (3*960) +1 = 0 + 1 = 1

Right Side: sec(3*960)= 1

sorry haley, that is wrong

tanx + 1 is not equal to secx

rather

tan^2 x + 1 = sec^2 x

see my solution below.

I forgot to include the following:

Since "squaring" took place in my solution, all answers should have been verfied.

Upon checking, we find that 30º does not work, since tan90º is undefined.

So all periodic answers based on 30º do not work

sin (u) + cos (u) = 1 --->

sqrt(2)sin(u+pi/4) = 1

Note that

sin(a+b) = sin(a)cos(b) + cos(a) sin(b)

You can use this rule to write a sum of sin and cos as a single sin or cos.

sin(u+pi/4) = 1/sqrt[2] --->

u = 0 Mod(2 pi) OR u = pi/2 Mod(2 pi)

In this case you could have found the two solutions u = 0 and

u = pi/2 by inspection. Because there can be only two solutions in an interval of 2 pi, you then know that these are all the solutions in such an interval. All other solutions differ from these by a multiple of 2 pi.