Prep Calc Exam (2)

[CrystalTennis]

cos²x + sin²x=1 I substituted and got (1-sin²x) +sinx =1
then i simplified to sin²x-sinx= 0
and now i'm stumped and I am supposed to solve where 0 is less than or equal to x and x is less than 2pi

 

[Yogi]

cos²x + sin²x=1 I substituted and got (1-sin²x) +sinx =1 what happened to the 2?
then i simplified to sin²x-sinx= 0
and now i'm stumped and I am supposed to solve where 0 is less than or equal to x and x is less than 2pi

Draw a triangle with angle x and label adjacent, opposite and hypotenuse.
cosx=A/H
sinx=O/H

Therefore:
\(\displaystyle
\frac{A^2}{H^2} + \frac{O^2}{H^2} = 1\)
........
........omitting steps for speed since you can fill in the rest (use pythagorean theorem)
1=1

This is true no matter what values x takes so your solution will be \(\displaystyle 0 \leq x< 2 \pi\)

This is one of those common identities that you should just have committed to memory.

 

[CrystalTennis]

the question was to solve cos²x+sinx=1 the sinx isn't sin²x
I think i'm supposed to simplify to use a unit circle
okay so sin²x-sinx=2 what do i do from here?

 

[Yogi]

(1-sin²x) +sinx = 1

-sin²x+sinx = 0
sin²x-sinx = 0
sinx(sinx-1) = 0

What values of sinx makes this true?
Should be 4 solutions.

 

[CrystalTennis]

okay so if i split it up sinx=0 and sinx-1=0
the values that make this true are 0pi, pi/2, pi, and 2pi

 

[Yogi]

okay so if i split it up sinx=0 and sinx-1=0
the values that make this true are 0pi, pi/2, pi, and 2pi

0pi is just written as 0. If you start at 0 and go around 2pi radians you end at 0, so 0 and 2pi are equivalent. Also, \(\displaystyle 2 \pi\) doesn't fall in the given possible values for x: \(\displaystyle 0 \leq x< 2 \pi\)

So you have 3 of them correct.