Question

Find the general solution:
Problem in the comments, will FAN and MEDAL

Answer 1

\[2\cos^2 x=2+sinx\]

Answer 2

HI!!

Answer 3

rewrite the left hand side a
\[2(1-\sin^2(x)\] then you will have a quadratic equation in sine

Answer 4

Kay, I wrote that

Answer 5

then multiply out

Answer 6

and set it equal to zero

Answer 7

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

Answer 8

wouldnt it be a plus on the second line?

Answer 9

yes i guess it would wouldn't it

Answer 10

\[2\sin^2(x)+\sin(x)=0\\
\sin(x)(2\sin(x)+1)=0\] that's better

Answer 11

how'd you get that second line? @misty1212

Answer 12

factored out the sine

Answer 13

what's next

Answer 14

set each factor equal to zero and solve

Answer 15

sinx=0 and sinx=-1/2?

Answer 16

yes

Answer 17

then solve for x

Answer 18

Do you know how I would write the solutions in the general form format? @misty1212

Answer 19

\[\sin(x)=0,x=0,\pi,2\pi,3\pi,...\] or
\[x=k\pi, k\in \mathbb{Z}\]

Answer 20

\[\sin(x)=-\frac{1}{2}\]
\[x=\frac{7\pi}{6}+2k\pi,x=\frac{11\pi}{6}+2k\pi\]

Answer 21

Thank you both @satellite73 and @misty1212

Answer 22

yw

Answer 23

\[\color\magenta\heartsuit\]