View Full Version : Trigonometry equation
lucky_girl
Jul 3, 2009, 05:31 PM
Solve each equation for principle angle values (where reasonable) or principle
angle inverse trigonometric functions at the proper point
1.2cos²(x)+cos(x)-1=0
2.sin(3x)=cos(3x)
Perito
Jul 3, 2009, 05:56 PM
Solve each equation for principle angle values (where reasonable) or principle
angle inverse trigonometric functions at the proper point
1.2cos²(x)+cos(x)-1=0
2.sin(3x)=cos(3x)
1. This is solved just like an ordinary equation. Try a simple substitution, Y = cos(x).
2cos^2(x)+cos(x)-1=0
becomes
2Y^2 + Y -1=0
You can solve this using the quadratic equation:
Y=\frac {-1 \pm \sqrt {5}}{2}
And substitute back into the equation
cos(x) = \frac {-1 \pm \sqrt {5}}{2}
Then
x = cos^{-1}\left(\frac {-1 \pm \sqrt {5}}{2} \right)
2.
sin(3x)=cos(3x)
\frac {sin(3x)} {cos(3x)} = 1 = tan(3x)
x = \frac {tan^{-1}(1)}{3}
lucky_girl
Jul 3, 2009, 09:49 PM
thank you so much...
I did 2nd one like this
sin3(0)=cos3(0)
tan3(0)=1
3(0)=45 degrees
0 = 15 degrees
am I right?
Perito
Jul 4, 2009, 04:41 AM
i did 2nd one like this
sin3(0)=cos3(0)
tan3(0)=1
3(0)=45 degrees
0 = 15 degrees
am i right?
This doesn't look like the second one you gave me. You had written "sin(3x)", including the parentheses. Now you're writing Sin3(0). Where did the "0" come from and is this Sin^3(0) or is it 3 Sin(0)? I'll assume the "0" is a typo and that the 3 comes inside the parentheses:
sin(3x) = cos(3x)
\frac {sin(3x)}{cos(3x)} = tan(3x) = 1
3x = tan^{-1}(1) = 45\, degrees
x = 15
Then you are correct.
Unknown008
Jul 4, 2009, 08:46 AM
LOL! I reckon the OP mean theta \theta
Perito
Jul 4, 2009, 09:05 AM
Some of these things zoom over my head.