The idea is that implies for all x, but not vice versa. The correct equivalence is
.
Thus, if you solve and find that, say,
,
it does not follow that
.
At most you can guarantee that
.
Indeed, in your first example
.
Dear all,
I'm solving 2 sums
a) 1 - cos x = (sqrt 3) sin x
By squaring both sides, and using the sin^2x + cos^2x = 1 relation,
I get 2 cos^2x - cos x - 1=0
Solving, I get x = 120, 240 (for 2nd, 3rd quadrant) and 0, 360 (for 1st, 4th quadrant)
May I know why 240 is rejected?
b) 2 + sin x = 4 cos x
Similarly, I'm solving down to 17 sin^2x + 4 sinx - 12 = 0
I get sin x = 0.7307 or -0.966
Hence, x - 46.9 or 133.1 ; 285 or 255
Why is 133.1 and 255 rejected?
Thanks
The idea is that implies for all x, but not vice versa. The correct equivalence is
.
Thus, if you solve and find that, say,
,
it does not follow that
.
At most you can guarantee that
.
Indeed, in your first example
.
Hello, Rodimus79!
;808717]Dear all,
I'm solving 2 sums
By squaring both sides, and using the sin^2x + cos^2x = 1 relation,
I get 2 cos^2x - cos x - 1=0
Solving, I get x = 120^{o}, 240^{o} (for 2nd, 3rd quadrant) and 0^{o}, 360^{o} (for 1st, 4th quadrant)
May I know why 240^{o} is rejected?
When squaring an equation, we must check our answers.
Some answers may not check out.
. .
. .
. .
. .
Hello again!
Do you know WHY squaring an equation can introduce "extra" solutions?
Consider the equation: .
The answer is, of course: .
Suppose we square the equation before solving.
Then we have: .
This is a quadratic equation; it will have two solutions.
. . .
We find that satisfies the equation,
. . but does not.
Therefore, is an extraneous solution.