Question 561795
find values of x for 6 sin^2(x) = 3 between [0,2pi]
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6 sin^2(x) = 3
Divide by 6:
sin^2(x) = 1/2 -> sin^2(x) - 1/2 = 0
Factor:
{{{(sin(x)+1/sqrt(2))(sin(x)-1/sqrt(2)) = 0}}}
This gives {{{sin(x) = 1/sqrt(2)}}}, {{{-1/sqrt(2)}}}
On the interval [0,2pi], the solutions are pi/4,3pi/4,5pi/4,7pi/4