$\displaystyle 4sin^2x-3=0$
How do I find the zeros in exact values?
Hello, senna!
You've never seen one of these before?
Just use your knowledge of algebra . . . then Trig at the end.Solve for $\displaystyle x\!:\;\;4\sin^2\!x-3\:=\:0$
Add 3 to both sides: .$\displaystyle 4\sin^2\!x \:=\:3$
Divide by 4: .$\displaystyle \sin^2\!x \:=\:\frac{3}{4}$
Take square roots: .$\displaystyle \sin x \:=\:\pm\frac{\sqrt{3}}{2}$
Can you finish it now?