set it equal to zero then you will get quadratic equation
Yeah I noticed that but can you show me how you to set it up
\[\huge\rm 7\sin^2x-14sinx+2+5=0\] add 5 both
to make it easy let sin(x) x \[\huge\rm 7x^2 -14x+2+5 =0\] |dw:1437508829146:dw| solve for x :-)
Ohh okay thanks
let me know what you get
Yeah just give me a min I need to solve another problem real quick
equal to zero 0 then take out common factor
opps sorry i didn't get notification
and yes that's right now factor this quadratic equation x^2 -2x +1
yes right now set these equal to zero \[\huge\rm 7 \cancel= 0 ~~~(x-1)=0~~~(x-1)=0\] solve for x
what about the sines though?
yes now change x with sin
remember we changed sin(x) with x right ?
so \[\huge\rm sinx =1 \]
now use unit circle to find solution
find where one is 1 is y on the unit circle?
(cos , sin) x coordinate represent cos function and y coordinate represent sin function
would it be pi/2
thats it, thats the only solution
oh okay thank you so much!
my pleasure but remember sinx = 1 multiplicity of 2
I'm gonna post another question can you help me with that one as well, I think i know the answer just confused how to get to the answer
so my answer would be sinx=1^2?
no sin x= 1 is right
so just to confirm, my answer box would read sinx=1