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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.