Helene is asked to solve the equation #3sin^2theta-2sintheta=0# for #0<=theta<=pi#, she find that #theta=pi#. How could she check whether this is a correct root for the equation?
1 Answer
Nov 24, 2016
see explanation.
Explanation:
Substitute the solution, that is
#theta=pi# into the original equation and if it makes the equation true then it is a correct root of the equation.
#" Check that" 3sin^2theta-2sintheta=0" is true"#
#rArr3sin^2(pi)-2sin(pi)#
#"Now" sin(pi)=0#
#rArr3(0)^2-2(0)=0larr" satisfies equation"#
#"thus " theta=pi" is a correct root"#
