Question

find all solutions for 10sin(x) ^2 - sin (x) = 2

Answer 1

\[10\sin^2(x)-\sin(x)-2=0\]
let u=sinx
so u^2=(sinx)^2
\[10u^2-u-2=0\]
\[10u^2-5u+4u-2=0\]
\[5u(2u-1)+2(2u-1)=0\]
(5u+2)(2u-1)=0
u=-2/5
sinx=-2/5
u=1/2
sinx=1/2
solve these for x

Answer 2

wowee! that was quick!

Answer 3

you are a factoring somebody! love how you split the middle term! i can't do it

Answer 4

of course you chickened out on how to solve
\[\sin(x)=-\frac{2}{5}\] for x!

Answer 5

satellite it is easy

Answer 6

ax^2+bx+c
just find two factors of a*c that add up to be b
say you found m*n=a*c and m+n=b
just replace bx with mx+nx and woolah!

Answer 7

sinx = -2/5, so x=inverse sin(-2/5), which is in the 3rd or fourth quadrant

so x = 180+23.6 degrees, or 360-23,6 degrees