Expert: Steve Holleran - 11/19/2007

Question
QUESTION: I am working on the following problems and need to know if I am on the right track.

Problem 1

Solve the equation (find all solutions)

3tan^4(x) - 10tan^2 x + 3 = 0

My work so far:
Substituting u for tan^2;

3u^2-10v+3=0
u= -b+/- square root of b^2-4ac divided by 2a therefore,
u= 10 =/- squareroot of 10^2-4(3)(3) divided by 2(3),
u= 3
tan^2 x = 3
tan x = sqrt 3
This is where I get stuck.

The next one is:

Solve the equation(find all solutions):

4 sin^4 (x) - 7 sin^2 x + 3 = 0

Beginning as above;
4u^2 (x) -7 v x + 3 = 0
u= 1 or 3/4, u= 1 checks and u =3/4 doesn't.
This is where I am stumped.

Thanks for your help.

ANSWER: Hi again, Autumn,

Your work is good for what you have.  Just a couple considerations:

1.  When you get to using the quadratic formula, you get two solutions:

   u = 3 or u = 1/3

so  tan^2 x = 3 --> tan x = +/- sqrt3

diagram a right triangle with opposite side to the angle = sqrt3 and adjacent side 1, so the hypotenuse is 2  (remember tan = opp/adj?) and you'll see the angle is pi/3

so for tan x = +sqrt3, you have x = pi/3 +/- k * pi

for tan x = - sqrt3, you get x = 2pi/3 +/- k * pi

then tan x = 1/sqrt 3 gives x = pi/6 +/- k * pi

and  tan x = -1/sqrt 3 gives x = 5pi/6 +/- k* pi



2.  In this one, both the values should work.  What you've done is correct.  Just keep going:

sin^2 x = 1 --> sin x = 1 or sin x = -1

for sin x = 1, x = pi/2 +/- 2k * pi

for sin x = -1, x = 3pi/2 +/- 2k * pi

when sin^2 x = 3/4, then sin x = +- sqrt3 / 2

for sin x = + sqrt3 / 2, you have x = pi/3 +/- 2k * pi and x = 2pi/3 +- 2k * pi

for sin x = - sqrt3 / 2, you have x = 4pi/3 +- 2k * pi and x
= 5pi/3 +- 2k * pi

I hope this is what you needed.

You do nice work.  just use the reference triangles to find the trig values.

Let me know if any of this is not clear.
Steve

---------- FOLLOW-UP ----------

QUESTION: Thank you,

Could you elaborate some on this part: "just use the reference triangles to find the trig values."  This is something that I seem to really not be grasping very well.


Answer
Hi Autumn,

Okay, when you want to find a trig value for "special" angles, those that are multiples of 30 or 45, you want to draw a coordinate axis system (x and y axes) and mark off the angle.

For example, lets take 30 degrees or pi/6.

Draw an angle with the vertex at the origin and the initial side along the positive x-axis.  then draw a 30 degree angle to the terminal side.  Now, if you drop an altitude from the terminal side to the x-axis, you have a right triangle-- the "reference triangle".  the reference angle is always the one to the x-axis, here its 30.

Label the sides with the ratios from the 30-60-90 triangle:

side opp 30 = 1
side opp 60 = sqrt3
hypotenuse  = 2

then, using the 30 and the SOH CAH TOA ratios, find the  trig values:

sin 30 = opp/hyp = 1/2

cos 30 = adj/hyp = sqrt3 / 2

tan 30 = opp/adj = 1/ sqrt3


Lets say the angle is in another quadrant, like theta = 4pi/3.

Now, 4pi/3 = 240 degrees, so we rotate an angle into the third quadrant, and drop the altitude back to the x-axis again, and I hope you'll see that the reference angle is 60 (240 - 180).  Now, using the reference triangle, you should have the side along the negative x-axis as -1, the hypotenuse as 2 and the altitude as -sqrt3, using the ratios.

Then sin 4pi/3 = opp/hyp = -sqrt3 / 2

    cos 4pi/3 = adj/hyp = - 1/2

    tan 4pi/3 = opp/adj = -sqrt3 / -1 = sqrt 3

You can do this for any of these special angles.

I hope you can understand this okay.

Steve