Question

4sin^2x-3=0 whats the general solution

Answer 1

is it a multiple choice ? @lucasbeauchemin

Answer 2

\[4 \sin ^{2}x-3=0\]
\[2\sin ^{2}x=\frac{ 3 }{ 2 }\]
\[1-\cos 2x=\frac{ 3 }{ 2 },\cos 2x=-\frac{ 1 }{ 2 }=-\cos \frac{ \pi }{3 }=\cos \left( \pi \pm \frac{ \pi }{ 3} \right)\]
\[=\cos \left( 2 n \pi+\pi \pm\frac{ \pi }{ 3 } \right) \]
\[2x=2 n \pi+\pi \pm \frac{ \pi }{ 3 },x=\frac{ \left( 2 n+1 \right)\pi }{ 2 }\pm \frac{ \pi }{ 6 }\]
where n is an integer

Answer 3

@lucasbeauchemin sorry mate ..i got it wrong somewhere..the above is right

Answer 4

i see now thanks