Patricia Said:

Pre-Calculus - Solution of Sets?

We Answered:

The U stands for union. It is any point that's in either one of the sets.
1. (-2, 1)

The upside down U stands for intersection. It's the points that are in both sets.
2. [0,6]

3. Since (-infinity, -2) and [0, infinity) don't intersect, this set is empty. The way to write an empty set is with a 0 crossed out.

Denise Said:

Pre Calculus: 2sin^2 X = sinX ... Find ALL solutions between (0,2pi)?

We Answered:

2sin²x = sinx
2sin²x - sinx = 0
sinx(2sinx - 1) = 0

sinx = 0, 1/2

x = ?/6, 5?/6, ?

Since the interval is open on both ends x = 0 and x = 2? are excluded.

Elsie Said:

What would be the general solutions to this Pre-Calculus math problem?

We Answered:

5 cos? - 2sin? = 3
Remember sin²? - cos²? = 1. Therefore,
sin? = ?[1 - cos²?].
Substituting,
5 cos? - 2 ?[1 - cos²?] = 3
To make life easier, let u = cos?. The equation becomes:
5u - 2?[1 - u²] = 3
Rearranging,
5u - 3 = 2?[1 - u²];
Squaring both sides,
(5u - 3)²=(2?[1 - u²])²;
25 u - 30u + 9 = 4(1+u²);
25u - 30u + 9 = 4 + 4u²;
29u² - 30u + 5 = 0
Using the quadratic formula we get u = 0.5994 and u = 1.360. Therefore,
cos? = 0.5994 and cos? = 1.360, and
? = 34.3° and ? = 77.9°
but remember cos? = cos (-?)
so 360° - 34.3° = 325.7° = ?
and 360° - 77.9° = 282.1° = ?
When solving equations with radicals, we must resubstitute all these four answers to check:
5cos34.3° - 2sin34.3° = 3.00
5cos77.9° - 2sin77.9° = 5.25
5cos325.7° - 2sin325.7° = 5.25
5cos282.1° - 2sin282.1° = 3.00
Therefore, only ? = 34.3° and 282.1° are answers.

Pretty tough question.

Cathy Said:

Pre Calculus Question: How many solutions does a linear equation in one variable have?

We Answered:

Usually only one.

Anna Said:

Pre-Calculus?

We Answered:

1)

sin(2x) + cos(2x)

=>2sinx cosx + cos^2(x) - sin^2(x)

=>2sinx cosx + (cosx + sinx)(cosx - sinx)

2)

cos(2x) + 2sinx = 0

1 - 2sin^2(x) + 2sinx = 0

2sin^2(x) - 2sinx - 1 = 0

sinx = [2 +/- sqrt(4+8)]/4

sinx = [2 +/-2sqrt(3)]/4

sinx = 1/2 +/- sqrt(3)/2

sinx = -0.366(ignoring invalid value of 1.36)

x = 201.5 , or 338.5 degrees in the interval [ 0 to 360 degrees]

3)

cos(2A) = 1 - 2sin^2(A)

sin^2(A) = [1 - cos(2A)]/2

sin^2(5pi/12) = (1/2) [1 - cos(5pi/6)]

=>(1/2) [1 - cos(pi-pi/6)]

=>(1/2) [1 + cos(pi/6)]

sin^2(A) = (1/2) [1 + ?3/2]

sin(A) = ±?[(1/2)(1 + ?3/2)]

Alberto Said:

Pre calculus?

We Answered:

21 for the first one.

?[?(x-5) + x] = 5
?(x-5) + x = 25
?(x-5) = 25 - x
x-5 = x^2 - 50x + 625
x^2 - 51x + 630 = 0
(x-21)(x-30) = 0
x = 21 or 30
But if you go back and substitute, you will find that 30 doesn't work, and this arises because we are only concerned with principal square roots in the equation. This type of thing will happen often in radical equations. There are really two square roots of every number, (e.g., (-5)^2 = 5^2 = 25), but ?25 is to be interpreted as 5 only.



3 for the second.

2x + ?(x+1) = 8
?(x+1) = 8 - 2x
x+1 = 4x^2 - 32x + 64
4x^2 - 33x + 63 = 0
(4x - 21)(x-3) = 0
x = 21/4 or 3
As above, we reject 21/4 as an extraneous root, since
2(21/4) + ?(21/4+1) = 21/2 + ?(25/4) = 21/2 + 5/2 = 26/2 = 13 and not 8.


4 or 9 for the third

Here's a different type of approach for this one (substitution):
Let y = ?x
then x = y^2
the equation becomes:

y^2 - 5y + 6 = 0
(y-3)(y-2) = 0
y = 2 or 3
and x = y^2 , so
x = 4 or 9
Both work this time.