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.
Respond to this Question
Similar Questions

Calculus  Integration
Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you! 1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx It says u = tan x to substitute So if I use u = tan x, then my du = secx^2 then I … 
calculus
find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x … 
math
1. (sinx/cscx)+(cosx/secx)=1 2. (1/sinxcosx)(cosx/sinx)=tanx 3. (1/1+cos s)=csc^2 scsc s cot s 4. (secx/secxtanx)=sec^2x+secxtanx 5. (cosx/secx1)(cosx/tan^2x)=cot^2x 
Calculus
find the derivative of f(x)=tanx4/secx I used the quotient rule and got (sec^2x4)(secx)(tanx4)(cosx)/sec^2x but I'm pretty sure that's wrong. Help. Thanks. 
precal
1/tanxsecx+ 1/tanx+secx=2tanx so this is what I did: =tanx+secx+tanxsecx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)(1/cosx) =sinx/cosx+ sinx /cosx= 2tanxI but I know this can't be correct because what I did doesn't end as a negatvie:( … 
precal PLEASE HELP!
secx/secxtanx=sec^2x+ secx+tanx 
Math
I can't find the integral for (tanx)^(6)*(secx)^(2) I tried splitting up tanx into (tanx)^2*(tanx)^4 and let the latter equal (secx)^2  1. Please help, thanks! 
Trig
Verify the identity. 1/(tanxsecx) + 1/(tanx+secx) = 2tanx 
Trig
Verify the identity. (secx + tanx)/(secx  tanx) = (1 + 2sinx + sin(^2)x)/cos(^2)x 
Calculus
Int tanx sec^2x dx can be taken as (by putting it in form of Int xdx) Int secx.secx tanx dx=(sec^2x)/2 Int tanx.sec^2x dx=(tan^2)/2. Which one is correct and why is the difference?