Expert: Abe Mantell - 1/1/2007

Question
I have two trig questions to ask.

13)
solve for x between 0 and 2pi
2sin squared x = 2+ cos2x

i have answers: pi/3, 2pi/3, 5pi/6, and 4pi/3 are these right?

15) Solve for x between 0 and 360 degrees

3 sin(x+20)=-1

sin (x+20)=-1/3
x+20= inverse sin (-1/3)= 19.47122063
x= -20-19.47122063 x=-39.47122063  
360-39.47122063  = 320.5287794
rounds to 320.5 degrees is this right?
please help me..

Please tell me if this is right, complete, and exact/simplest form. Thanks.


Answer
Hello Sara,
13)
First, convert sin^2 (x) to [1-cos(2x)]/2, to get:
2(1-cos(2x))/2 = 2 + cos(2x), which simplies to
cos(2x)=-1/2...
so 2x=2PI/3 or 2x=4PI/3 or 2x=8PI/3 or 2x=10PI/3
Thus, x= PI/3 or 2PI/3 or 4PI/3 or 5PI/3

15)
320.47122 is one answer...the other one is x+20=180+19.47122
which gives the other answer: x=179.47122

OK?

Happy New Year!

Abe