Question 806108
What are the values of t on the interval between [0,2pi] for the square root of 6 sin 2t + 3 tan 2t=0? Only the "6" is in a square root.
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I assume 2t is ^2(t)
..
{{{sqrt(6)sin^2(t)+3tan^2(t)=0}}}
{{{sqrt(6)6sin^2(t)+3sin^2(t)/cos^2(t)=0}}}
{{{sin^2(t)(sqrt(6)+3/cos^2(t))=0}}}
sin^2(t)=0
sin(t)=0
t=0, π
or
√6+3/cos^2(t)=0
3/cos^2(t)=-√6
cos^2(t)=3/-√6
reject(can't take sqrt of negative number)