solving trig. equations

posted by .

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 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

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

First Name
School Subject
Your Answer

Similar Questions

  1. 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 …
  2. 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 …
  3. math

    1. (sinx/cscx)+(cosx/secx)=1 2. (1/sinxcosx)-(cosx/sinx)=tanx 3. (1/1+cos s)=csc^2 s-csc s cot s 4. (secx/secx-tanx)=sec^2x+secxtanx 5. (cosx/secx-1)-(cosx/tan^2x)=cot^2x
  4. Calculus

    find the derivative of f(x)=tanx-4/secx I used the quotient rule and got (sec^2x-4)(secx)-(tanx-4)(cosx)/sec^2x but I'm pretty sure that's wrong. Help. Thanks.
  5. precal

    1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-secx =(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:( …
  6. precal- PLEASE HELP!

    secx/secx-tanx=sec^2x+ secx+tanx
  7. 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!
  8. Trig

    Verify the identity. 1/(tanx-secx) + 1/(tanx+secx) = -2tanx
  9. Trig

    Verify the identity. (secx + tanx)/(secx - tanx) = (1 + 2sinx + sin(^2)x)/cos(^2)x
  10. 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?

More Similar Questions