## anonymous one year ago tanA+sqart3=0,sinA-1/2=0,cosA+squrt3/2,secA-1=0 . which one does not have solution in qurdrant 2 ?

1. anonymous

i know i have to use the cast rule but still confused

2. Michele_Laino

first equation: $\tan A = - \sqrt 3$ we have the subsequent drawing: |dw:1434998122110:dw|

3. Michele_Laino

as we can see one solution belongs to the II quadrant, and the oter one belongs to the fourth quadrant

4. Michele_Laino

secon equation: $\sin A = \frac{1}{2}$ |dw:1434998289606:dw| here we have one solution which belongs to the first quadrant, and the second one which belongs to the second quadrant

5. anonymous

what about the cast rule ?

6. Michele_Laino

third equation: $\cos A = - \frac{{\sqrt 3 }}{2}$ |dw:1434998435741:dw| here we have one solution which belongs to the II quadrant, and the second one which belongs to the III quadrant

7. Michele_Laino

finally, fourth equation: $\sec A = 1 \to \frac{1}{{\cos A}} = 1 \to \cos A = 1$ |dw:1434998618085:dw| here we have 2 solutions, namely the first one which is A=0, and then it belongs to the I quadrant, whereas the second one, which is A=360, belongs to the IV quadrant. So what can you conclude?

8. anonymous

do u know how to do this without using the unit circle ?

9. anonymous

my teacher did the cast rule and it was much faster , kinda forgot though

10. Michele_Laino

yes! I know, we have to solve all of your four equations. Nevertheless I think that a graphical solution is the best solution

11. Michele_Laino

sorry, what is the cast rule?

12. anonymous

btw do u think the solution is thre last one ?

13. Michele_Laino

yes! That's right!

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