Sunday

August 31, 2014

August 31, 2014

Posted by **Emily** on Sunday, June 21, 2009 at 8:55pm.

1.) 3sin(x)+1=0, x within [0,2pi)

2.) 2sin(sq'd)(x)+cos(x)-1=0, x within R

3.) 4sin(sq'd)(x)-4sin(x)-1=0, x within R

4.) sin(x)+1=cos(x), x within [0, 2pi) -check for extraneous solutions.

- trig -
**bobpursley**, Sunday, June 21, 2009 at 9:07pmon 2 remember that sin^2 x = 1-cos^2 x, then use the quadratic formula,or factor.

on 3, factor

on 4,try squaring both sides, then substitue to get rid of the cos^2

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