Question

Find all solutions to the equation.

7 sin^2x - 14 sin x + 2 = -5

Answer 1

HI!!

Answer 2

let sin x=t

Answer 3

solve \[7u^2-14u-7=0\]

Answer 4

hello

Answer 5

then replace \(u\) by \(\sin(x)\)

Answer 6

7t^2-14t+7=0
t^2-2t+1=0

Answer 7

solve it as a regular quadratic equation

Answer 8

(t-1)^2=0
t=+1,-1

Answer 9

case 1: sin x=1
case 2: sin x=-1

Answer 10

yeah i made a mistake, should be \[7u^2-14u+7=0\] but the above answer is incorrect

Answer 11

so what would the solutions be?

Answer 12

oh sorry sin x=-1 cant be possible
only sin x=1

Answer 13

\[(u-1)^2=0\\
u=1\\
\sin(x)=1\] etc

Answer 14

those all the solutions?

Answer 15

\[\sin(x)=\pm1\]
CASE 1
\[\sin(x)=1\]
\[x=(4n+1)\frac{\pi}{2}, n \in \mathbb{Z}\]

Answer 16

similarly solve for sin(x)=-1

Answer 17

huh?

Answer 18

You have to solve for x,
when sin(x)=1,
x can take the values \[\frac{\pi}{2},\frac{5\pi}{2},\frac{9\pi}{2},.....\]\[\frac{\pi}{2},2\pi+\frac{\pi}{2},4\pi+\frac{\pi}{2},.....\]\[(2\times 0 \times \pi)+\frac{\pi}{2},(2 \times 1 \times \pi)+\frac{\pi}{2},(2 \times 2 \times \pi)+\frac{\pi}{2}\]\[2n \pi+\frac{\pi}{2},n \in \mathbb{Z}\]Thus there are infinite solutions, for any integral value of n\[-3,-2,-1,0,1,2,3\]

Answer 19

which is the same as \[(4n+1)\frac{\pi}{2}=2n \pi+\frac{\pi}{2}\]

Answer 20

@ganeshie8