yr11 2U trig

**deej813** · August 26th 2009, 10:25 PM

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

**Chris L T521** · August 26th 2009, 10:35 PM

Originally Posted by deej813

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

Applying the zero product property, we have either

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

Can you take it from here?

**deej813** · August 26th 2009, 10:50 PM

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

**Chris L T521** · August 26th 2009, 11:09 PM

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

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

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

Does this make sense?

**deej813** · August 26th 2009, 11:15 PM

yes i get it
finally

thanks

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