3sin^2(x) - 2cos(x) - 3 = 0, for 0 < x < 180.
3(1-cos^2(x)) - 2cos(x) - 3 = 0
3 - 3cos^2(x) 2cos(x) - 3 = 0
What happened to the "minus" sign in front of the "2cos(x)"?
Where did the "minus" sign come from? (This sort of algebra "magic" leads to serious errors once one arrives at more-complicated contexts, such as trigonometry. It can also lead to zero points for any work past the "magic" step.)
-3cos(x) = 0 or cos(x) = 0
The last line above implies that you factored the quadratic form -3Y
2 - 2Y to get (-3Y)(Y). How did this "factoring" happen? In particular, where went the subtraction? :shock:
Instead, try using what you learned back in algebra. Start with the equation:
. . . . .\(\displaystyle -3Y^2\, -\, 2Y\, =\, 0\)
Multiply through by -1 (or, which results in the same thing, add the terms to the other side of the equation) to get nicer signs:
. . . . .\(\displaystyle 3Y^2\, +\, 2Y\, =\, 0\)
Then do the
simple factoring:
. . . . .\(\displaystyle Y(3Y\, +\, 2)\, =\, 0\)
Then set each
factor (rather than what appears to be each
term) equal to zero, and solve.