# Trigonometric equation

• Jul 17th 2010, 03:04 PM
Angie80
Trigonometric equation
Can anybody help me with this equation? I am staring to think there is a mistake there....
I tried to factor it, but it does not seem right...

2sin^2(x) - 6sin(x)- 3=0
• Jul 17th 2010, 03:11 PM
Also sprach Zarathustra
Hint:

• Jul 17th 2010, 03:23 PM
e^(i*pi)
Hint 2:

$|sin(x)| \leq 1$
• Jul 17th 2010, 03:36 PM
Angie80
This is what I tried:
sinx=t
2t^2-6t-3=o
I used quadratic formula, and got:
t= ( 3- √15) /2 = -0.43

What now? I could substitute t for sinx now, but it does not help...
• Jul 17th 2010, 04:18 PM
mr fantastic
Quote:

Originally Posted by Angie80
This is what I tried:
sinx=t
2t^2-6t-3=o
I used quadratic formula, and got:
t= ( 3- √15) /2 = -0.43

What now? I could substitute t for sinx now, but it does not help...

Exact solutions:

Let $x_1 = \frac{3 - \sqrt{15}}{2}$. Then $\sin (x) = x_1 \Rightarrow x = \sin^{-1} (x_1) + 2 n \pi$ or $x = \sin^{-1} (-x_1) + \pi + 2 n \pi$ (using the symmetry of the unit circle) where n is an integer.

Approximate solutions:

Use a calculator to get a decimal approximation of $\sin^{-1} \left( \frac{3 - \sqrt{15}}{2} \right)$ and then proceed in the usual way.
• Jul 17th 2010, 04:50 PM
Angie80
So, I calculated it and got appx: -26
I must be missing something, because I really do not know how to proceed from here... :(
I know how to use unit circle, but what is value of x, when sinx=-26??

I am veeery confused now....
• Jul 17th 2010, 06:49 PM
mr fantastic
Quote:

Originally Posted by Angie80
So, I calculated it and got appx: -26
I must be missing something, because I really do not know how to proceed from here... :(
I know how to use unit circle, but what is value of x, when sinx=-26?? Mr F says: No. x = -26, not sinx.

I am veeery confused now....

sin is negative in the 3rd and 4th quadrants. Using degrees, in the 4th quadrant, x = -26 + (360)n. In the 3rd quadrant, x = (180 + 26) + (360)n by symmetry of the unit circle.

Go back and review examples from your class notes and textbook.
• Jul 17th 2010, 06:52 PM
Angie80
Thanks soooo much!