Question

solve 2sin^2X=4sinX+6

Answer 1

2sin^2X=4sinX+6
\[2 \sin 2x = 4 \sin x + 6\]

it looks like this, isn't it?

Answer 2

It is 2sin to the power of 2 x = 4sinx+6

Answer 3

ahhh

Answer 4

\[2 (\sin x )^2 = 4 \sin x + 6\]

Answer 5

like this ?

Answer 6

we assume sin x to be Y
so we have

\[2 Y^2 = 4Y + 6\]

Answer 7

get it?

Answer 8

\[2Y^2 = 4Y + 6\]
\[2Y^2 - 4Y - 6 = 0\]
\[y^2 - 2Y - 3 = 0\]
\[(Y+1) (Y-3) = 0\]

Answer 9

\[2\sin ^{2}x=4sinx+6\]

Answer 10

get it ?

Answer 11

it is like this \[2\sin ^{2}x=4sinx+6\]

Answer 12

yess then we assume sin x to be Y like what i said before

Answer 13

\[Y = -1\]
because of Y equal to sin x, then
\[\sin x = -1\]
x = \[x = \sin^{-1} (-1) = -\frac{ \pi }{ 2 } \]

Answer 14

let us prove it

Answer 15

We have
\[X = -\frac{ \pi }{ 2 }\]
substitute this X to
\[2 \sin^{2}x = 4 \sin x + 6\]
then we have
\[2 \sin^{2}(-\frac{ \pi }{ 2 }) = 4 \sin (-\frac{ \pi }{ 2 }) + 6\]
2 (-1)^2 = 4 (-1) + 6
2 = -4 + 6
2 = 2

Of course we have proved it! :)