Question

solve each equation on the interval (0,2pie) use exact values where possible or give solutions correct to four decimal places

4sin^2x=1

Answer 1

start with \[\sin^2(x)=\frac{1}{4}\] and then solve \[sin(x)=\pm\frac{1}{2}\]

Answer 2

what about cos2x-sinx=1

Answer 3

I will post the solution in a sec

Answer 4

Rohangrr, please allow me to post the complete solution

Answer 5

and perhaps even a correct one

Answer 6

It will be correct smartie

Answer 7

okay! @Hero

Answer 8

lol i have more questions

Answer 9

@satellite73 , I know I make mistakes, but I usually correct them.

Answer 10

i am sure it will be, i was referring to the fact that the last two rohangrr answers were random
i am sure you (hero) will have no trouble with this

Answer 11

\[\cos^2x - \sin x = 1\]\[(1 - \sin^2x) - \sin x = 1\]\[1 - \sin^2 x - \sin x = 1\]\[-\sin^2x - \sin x = 1 - 1\]\[-\sin^2x - \sin x = 0\]\[-(\sin^2x + \sin x) = 0\]\[\sin^2x + \sin x = 0\]\[\sin x(\sin x + 1) = 0\]

Now use zero product property to continue solving for x

Answer 12

sin x = 0
sin x + 1 = 0

Answer 13

thank u