SOLUTION: Notes:^-exponent problem: Cos2@+6Sin^2@=4 So far I have: 1-2Sin@+6Sn^2@=4 -2Sin@+6Sin^2@=3 4Sin^2@/4=3/4 (divide by 4) Sin^2@=3/4 Sin=+/-radical 3/4 = radical 3 over -2

Sin^2@=3/4
Sin=+/-radical 3/4
= radical 3 over -2

How is +/-radical 3/4
changed to radical 3 over -2 
Thank You
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You are confused about the symbol ±.  It means

"positive or negative" 

It's written ± 

(hold down alt while typing 0177 on the keypad with Num Lock on.) 

±5 means either one of the two numbers +5 or -5.

±23.684 means either one of thes two numbers +23.684 or -23.684

It's not +/-, but sometimes when people don't know the alt symbols

they write +/- for ±

You mistakenly took the forward slash / to mean "divided by".  It

doesn't mean "positive divided by negative"

Let's go thru the problem:

cos(2x) + 6·sin²(x) = 4

1 - 2·sin²x + 6·sin²(x) = 4

4·sin²(x) = 3

sin²(x) = 3/4

Right here is where you take square roots of 
both sides.  Every number has two square roots.
So 3/4 has two square roots  +Ö3/2 and -Ö3/2.
           _
sin(x) = ±Ö3/2

That represents two equations:
           _                     _
sin(x) = +Ö3/2   and  sin(x) = -Ö3/2

so your answer will be all angles with 
reference angle 60° or p/3 radians taken
in every quadrant.

Edwin
AnlytcPhil@aol.com