Equation solutions

• Dec 11th 2009, 08:44 AM
Stuck Man
Equation solutions
I want to find x-axis intercepts of this equation:
(sin x+0.5)(sin x-0.25)=0.25

0 <= x <= 360.

So sin x=-0.25 or sin x=0.5

One of the associated acute angles is 30 which is found from sin x=0.5. The other is 48.59 but I don't know where it comes from.
• Dec 11th 2009, 09:29 AM
Raoh
hi(Hi)
put $\displaystyle \sin x=X$
and solve $\displaystyle (X+0.5)(X-0.25)-0.25=0$
• Dec 11th 2009, 09:29 AM
Stuck Man
I rearranged the equation to equal 0 using +.75 and -.5. I then solved the problem but I don't know why I couldn't solve it with the other equation.
• Dec 11th 2009, 09:35 AM
Stuck Man
I think it must be necessary to factorise without any added on value. Completing the square for instance wouldn't work.
• Dec 11th 2009, 09:48 AM
Raoh
Quote:

Originally Posted by Stuck Man
I think it must be necessary to factorise without any added on value. Completing the square for instance wouldn't work.

(Thinking)
to find the x-intercepts you must solve the given equation,

$\displaystyle (X+0.5)(X-0.25)-0.25=X^2-0.25X+0.5X-0.375=X^2+0.25X-0.375=0$.
i found the solutions $\displaystyle 0.5$ and $\displaystyle -0.75$
therefore,
$\displaystyle \sin x=0.50 \Leftrightarrow x=\arcsin (0.50)$
or
$\displaystyle \sin x=-0.75\Leftrightarrow x=\arcsin (-0.75)$.
• Dec 11th 2009, 09:59 AM
Stuck Man
I think you've mixed up the signs of the solutions. You've rewritten the formula the same as I did. I'm still not sure why I have to fully factorise and not use the Completing The square method. I think it have something to do with the use of sin.
• Dec 11th 2009, 02:21 PM
skeeter
Quote:

Originally Posted by Stuck Man
I want to find x-axis intercepts of this equation:
(sin x+0.5)(sin x-0.25)=0.25

0 <= x <= 360.

So sin x=-0.25 or sin x=0.5

One of the associated acute angles is 30 which is found from sin x=0.5. The other is 48.59 but I don't know where it comes from.

$\displaystyle \sin^2{x} + .25\sin{x} - 0.125 = 0.25$

$\displaystyle \sin^2{x} + .25\sin{x} - 0.375 = 0$

$\displaystyle 8\sin^2{x} + 2\sin{x} - 3 = 0$

$\displaystyle (4\sin{x} + 3)(2\sin{x} - 1) = 0$

$\displaystyle \sin{x} = \frac{3}{4}$ ...

$\displaystyle x = \arcsin\left(\frac{3}{4}\right) \approx 48.59^\circ$

$\displaystyle x = 180 - \arcsin\left(\frac{3}{4}\right) \approx 131.4^\circ$

$\displaystyle \sin{x} = \frac{1}{2}$

$\displaystyle x = 30^\circ$

$\displaystyle x = 150^\circ$
• Dec 11th 2009, 10:11 PM
Stuck Man
I still don't know why I can't use the equation as it was originally:
(sin x+0.5)(sin x-0.25)=0.25

sin x=-0.25 or sin x=0.5
• Dec 11th 2009, 11:11 PM
11rdc11
Quote:

Originally Posted by Stuck Man
I still don't know why I can't use the equation as it was originally:
(sin x+0.5)(sin x-0.25)=0.25

sin x=-0.25 or sin x=0.5

You can't do that in math.

Now if the equation was

$\displaystyle (\sin{x}+.5)(\sin{x}-.25) = 0$

then you could do that

Notice how Raoh and Skeeter set the equation equal to 0 and then solved.