May 26, 2016

Homework Help: solving trig. equations

Posted by Jen on Thursday, April 12, 2007 at 5:34pm.

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


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 dont 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 = 0,180,360
or 0, pi/2, pi in radians

so x = 0, 30,60 or 0,pi/6,pi/3

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

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

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

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 30 does not work, since tan90 is undefined.
So all periodic answers based on 30 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.

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