Without solving, how do you determine the number of solutions for the equation #sintheta=sqrt3/2# in #0<=theta<2pi#?

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Without solving, how do you determine the number of solutions for the equation #sintheta=sqrt3/2# in #0<=theta<2pi#?

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Answer 1 · 2016-11-06T14:48:23

2 solutions
#theta=pi/3# and #theta=2pi/3#

Explanation:

#pi/3# is a remarkable angle for the sinus function and #sin(pi/3)=root2(3)/2#

on the other hand we know that #sin(pi-theta)=sin(pi)cos(theta)-cos(pi)sin(theta)=sin(theta)#. This implies that if there is a solution at #theta=pi/3#, there must be another solution at #theta=2pi/3#

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Without solving, how do you determine the number of solutions for the equation #sintheta=sqrt3/2# in #0<=theta<2pi#?

1 Answer

Explanation:

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