# Solving Trig functions?

• March 27th 2011, 03:53 PM
dondonlouie
Solving Trig functions?
I'm not sure how to solve these trig functions;

1. 2sin(theta)−sqrt(2)tan(theta)= 0 ; 0 ≤ theta < 2pi

2. sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 ; 0 ≤ theta < 2pi
for #2 i factored out the sin(theta) (cos(theta)-1)- cos(theta)+1 =0
then moved the - cos(theta)+1 to the right side and solved for sin(theta)
is that right so far? D:
• March 27th 2011, 04:06 PM
e^(i*pi)
1. Multiply through by $]\cos(\theta)$ and solve by factoring
• March 27th 2011, 05:09 PM
topsquark
Quote:

Originally Posted by dondonlouie
2. sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 ; 0 ≤ theta < 2pi
for #2 i factored out the sin(theta) (cos(theta)-1)- cos(theta)+1 =0
then moved the - cos(theta)+1 to the right side and solved for sin(theta)
is that right so far? D:

It's probably better to rewrite that first line as
$sin(\theta) \cdot (cos(\theta) - 1) - (cos(\theta) - 1) = 0$

Can you see where to go from here?

-Dan
• March 27th 2011, 06:04 PM
dondonlouie
Quote:

Originally Posted by topsquark
It's probably better to rewrite that first line as
$sin(\theta) \cdot (cos(\theta) - 1) - (cos(\theta) - 1) = 0$

Can you see where to go from here?

-Dan

wouldn't the cos(theta)-1 cancel out leaving sin(theta)=0?
• March 27th 2011, 06:19 PM
topsquark
Quote:

Originally Posted by dondonlouie
wouldn't the cos(theta)-1 cancel out leaving sin(theta)=0?

Careful! When you have the equation ab = 0 then either a = 0 or b = 0. So what this equation is saying:
$(sin(\theta) - 1)(cos(\theta) - 1) = 0 \implies sin(\theta) - 1 = 0 \text{ or } cos(\theta) - 1 = 0$

-Dan
• March 27th 2011, 09:18 PM
dondonlouie
Quote:

Originally Posted by topsquark
Careful! When you have the equation ab = 0 then either a = 0 or b = 0. So what this equation is saying:
$(sin(\theta) - 1)(cos(\theta) - 1) = 0 \implies sin(\theta) - 1 = 0 \text{ or } cos(\theta) - 1 = 0$

-Dan

how did you get from sin(theta)(cos(theta)-1)-(cos(theta)-1)= 0
to (sin(theta)-1)(cos(theta)-1)?
• March 27th 2011, 09:35 PM
Prove It
Take out $\displaystyle (\cos{\theta} - 1)$ as a factor...
• March 27th 2011, 10:14 PM
dondonlouie
Quote:

Originally Posted by Prove It
Take out $\displaystyle (\cos{\theta} - 1)$ as a factor...

ohhh okay i see it :D
thank you so much for helping me :D
• March 28th 2011, 02:07 PM
topsquark
Quote:

Originally Posted by dondonlouie
ohhh okay i see it :D