Question

Solve for (0,2pi):
2cos^2x+3sinx-3=0

Answer 1

Trigonometric equations suck .. please help .. :(

Answer 2

plz c my attachment

Answer 3

Thank you! such great help! :) much appreciated! @shamim

Answer 4

u r most welcome

Answer 5

one more thing is
sin x=sin(π/6)=sin(π-π/6)=sin(5π/6)
so x=5π/6

Answer 6

if u r pleased on me then u can give me medal

Answer 7

just one question @shamim when you factored out the negative where did it go? it just disappeare on the 4th step?

Answer 8

-3+2=-1

Answer 9

i used it in my 4th step

Answer 10

still not clear? tell me plz

Answer 11

yea first you factored out the negative in -2sin^2x+3sinx-1=0 and made it -(2sin^2x-3sinx+1)=0 and then when you rewrote it you put 2sin^2x-3sinx+1=0. Where did the factored out negative go?

Answer 12

@shamim

Answer 13

i multiplied by a negative both side

Answer 14

u know -0=0

Answer 15

ohh okay that i get it :) thanks!

Answer 16

u r most welcome