Question

sin^2x - 2sinx + 1 = 0

Answer 1

Solve for x??

Answer 2

First solve for sin x, then go to the unit circle and find anywhere where sin equals x

Answer 3

this is an equation that is reducible to a quadratic.

its actually a perfect square

(sin(x) - 1)^2 = 0

now solve for x

Answer 4

^^ if you wanna solve for sinx from the above answer \[(sinx-1)^{2}\]=0
then introduce a squareroot in both sides to get : \[\sin x-1=0\]

Answer 5

Do you get what I am saying above?

Answer 6

So would you then add 1 to both sides, get sinx = 1, then according to the unit circle the answer would be pi/2 ?

Answer 7

it's for general solutions so actually pi/2 + 2pi n

Answer 8

I think that so

Answer 9

Yea you sinx will surely be equal to 1(one)

Answer 10

so (sinx - 1)^2 = 0 when worked out equals sin x - 1?

Answer 11

\[(sinx-1)^{2}=0 \]
\[\sqrt{(sinx-1)^{2}}=\sqrt{0}\]
\[(sinx-1)^{2*1/2} =0\]
\[\sin x-1=0\]
and therefore \[\sin x=1\]

Answer 12

I got sinx^2 + 1 when I worked it out