1. ## yr11 2U trig

for 0<= θ <=360, solve (sin θ +1)(2cos θ -1)=0

2. Originally Posted by deej813
for 0<= θ <=360, solve (sin θ +1)(2cos θ -1)=0
Applying the zero product property, we have either

$\displaystyle \sin\theta=-1$ or $\displaystyle \cos\theta=\tfrac{1}{2}$.

Can you take it from here?

3. do i just put sin^-1(-1) in my calculator
and cos^-1(1/2)
??
sorry im very confused

4. Originally Posted by deej813
do i just put sin^-1(-1) in my calculator
and cos^-1(1/2)
??
sorry im very confused
You could...but its best if you try to memorize these values from the unit circle: Unit circle - Wikipedia, the free encyclopedia

In our case, $\displaystyle \sin^{-1}\left(-1\right)$ has only one value in the specified interval for $\displaystyle \theta$, and its $\displaystyle \theta=270^{\circ}$

However, $\displaystyle \cos^{-1}\left(\tfrac{1}{2}\right)$ has two solutions in the given interval for $\displaystyle \theta$. They are $\displaystyle \theta_1=60^{\circ}$ and $\displaystyle \theta_2=300^{\circ}$.

Your final answer to the original question is all three values for $\displaystyle \theta$.

Does this make sense?

5. yes i get it
finally
thanks