sin (2x) = square root 3/2
interval - 0 less than, equal to x less than 2pi
What's the solution set......
i got pi/6 and 2pi/6 (is it right?)
What are more solution set??? I guess i need 4 of them.
thnxs :)
Interval [0, 2pi)
2 sin^2x-cosx-1=0
What is the solution set???
When i factored i got: -(2 sinx-1)(sinx+1)=0
sin x= -1/2 and -1
so isn't it solution set is: pi/3, 5pi/3, and 3pi/2???
Please help. thnxs :)
Does anyone know what are the controversies that is been going on since 1550's when maps were first made till today??? I need details so where can i find that...
Help needed thnxs!