SOLUTION: sorry, these problems aren't from a textbook. I've started these four problems then just started going in circles to find the answer. 1.)Solve for x, where x is between 0 and 2pie

Algebra ->  Trigonometry-basics -> SOLUTION: sorry, these problems aren't from a textbook. I've started these four problems then just started going in circles to find the answer. 1.)Solve for x, where x is between 0 and 2pie      Log On


   



Question 173155: sorry, these problems aren't from a textbook. I've started these four problems then just started going in circles to find the answer.
1.)Solve for x, where x is between 0 and 2pie
tan^(2)x-tanx-3=0
2.)verify the identity:
sinx/1-cos=cscx+cotx
where 0 3.)evaluate:
cos[sin^(-1)(3/5)]
4.) Given that sin theta =1/2, theta in Q1 and cos beta =1/2, find cos (theta-bata)
much appreciated!!


Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1.)Solve for x, where x is between 0 and 2pi
Tan%5E2x-Tan%28x%29-3=0

Write T for Tan%28x%29

T%5E2-T-3=0

That doesn't factor, so we use the quadratic formula:

T+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

with a=1, b=-1, and c=-3:

T+=+%28-%28-1%29+%2B-+sqrt%28%28-1%29%5E2-4%2A%281%29%2A%28-3%29+%29%29%2F%282%2A%281%29%29+

T+=+%281+%2B-+sqrt%281%2B12+%29%29%2F2+

T+=+%281+%2B-+sqrt%2813+%29%29%2F2+

Using the +

T+=+%281+%2B+sqrt%2813+%29%29%2F2+=+2.302775638

Replace T by Tan%28x%29

Tan%28x%29+=+2.302775638

Use inverse tangent key on calculator,
with calculator in radian mode:

x=1.161109817

However, the calculator only gives the
answer which is in quadrant I.  There is
also another answer in quadrant III, because
the tangent is also positive in quadrant III.
To get that, we add pi:

x=1.161109817%2Bpi=4.30270247

Using the minus:

Tan%28x%29+=+%281+-+sqrt%2813+%29%29%2F2+=+-1.302775638

Tan%28x%29+=+-1.302775638

Use inverse tangent key on calculator,
with calculator in radian mode:

x=-0.9161311535

However, the calculator only gives the
negative answer which is in quadrant IV,
taken as a negative angle. However to get
a coterminal angle between 0 and 2pi,
we add 2pi:

x=-0.9161311535%2B2pi=5.367054154
  
There is also another answer in quadrant II,
because the tangent is also negative in quadrant IV.
To get that, we add pi to the-0.9161311535

x=-0.9161311535%2Bpi=2.2254615


2.)verify the identity:




3.)evaluate:
matrix%281%2C5%2C%0D%0A%0D%0ACos%2C%22%28%22%2CSin%5E%28-1%29%2C3%2F5%2C%22%29%22+%29+

Let's draw the picture of matrix%281%2C2%2C%0D%0A%0D%0ASin%5E%28-1%29%2C3%2F5+%29+

That means the angle whose sine is 3%2F5

The inverse sine by definition is always between 
-pi%2F2 and pi%2F2. Thus the angle matrix%281%2C2%2C%0D%0ASin%5E%28-1%29%2C3%2F5+%29+ is in Quadrant I because 3%2F5
is a positive number.

Since the sine = y%2Fr we will let y be
the numerator of 3%2F5 and r be the
denominator.  

The angle matrix%281%2C2%2CSin%5E%28-1%29%2C3%2F5+%29+ is



Now we need the cosine of this angle, which is x%2Fr, so
we will need the value of x.  We use the Pythagorean theorem,

r%5E2=x%5E2%2By%5E2
5%5E2=x%5E2%2B3%5E2
25=x%5E2%2B9
25-9=x%5E2
16=x%5E2
sqrt%2816%29=sqrt%28x%5E2%29
4=x

So now we have x=4, so we put that in:



Therefore the cosine of that angle is x%2Fr or 4%2F5

So,




If I have time I'll come back to this:

4.) Given that sin theta =1/2, theta in Q1 and cos beta =1/2, find cos (theta-bata)


Edwin