solving trig. equations
posted by Jen .
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
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
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 --->
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.