trig
posted by saad .
Solve the equation for cos thetatan theta=0 for greater than or egual to zero but less than 2pi. Write your answer as a multiple of pi, if possible form the following choices:
pi , 5pi
A.  
4___ 4
B. pi 3pi 5pi 7pi
 ' ''
4_____ 4_____ 4______4
C. pi 3pi 3pi 7pi
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2_______4______2_____4
D. pi 7pi 3pi 11pi
 ' ''
2________6______2_____6

cos ( theta )  tan ( theta) = 0
tan ( theta ) = sin (theta ) / cos( theta )
cos ( theta )  tan ( theta ) =0
cos ( theta ) = tan ( theta )
cos ( theta ) = sin ( theta ) / cos ( theta )
cos^2 ( theta) = sin ( theta )
Now go on:
wolframalpha dot com
When page be open in rectangle type :
solve cos^2 ( theta) = sin ( theta )
and click =
After few seconds you will see solution.
Then click option Show steps 
cosØ  tanØ = 0
cosØ  sinØ/cosØ = 0
multiply by cosØ
cos^2Ø  sinØ = 0
(1  sin^2Ø)  sinØ = 0
sin^2Ø + sinØ  1 = 0
sinØ = (1 ± √5)/2
sinØ = .618034 or 1.608 , the last is not possible since sinØ has to be between 1 and +1
Ø = 38.17° from my calculator (or 141.83°)
All your choices are in radians in multiples of π/2, π/4 or π/6
which would be multiples of 90°, 45° or 30°
None of these match,
btw, my answers satisfies the original equation.