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)

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}

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?

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.

LOL! I reckon the OP mean theta \theta

Some of these things zoom over my head.