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there is a formula for sin(2x)
do you know what it is?
i dont think so
sin(2x) = 2sin(x)cos(x)
Equation: 7 sin2x - 14 sin x + 2 = -5 move 2 to the other side to get?
@q12157: Very nice start, very helpful. Be certain to let mabitrix find his or her own answer(s), with your guidance.
7 sin2x -14 sinx = -3
now plug in the formula for sin(2x) to get?
also its not -3 its -7
since you are subtracting 2 on both sides
7 2sin(x)cos(x) -14 sinx = -7 ?
7 2sin(x)cos(x) is the same as 14sin(x)cos(x) right?
aren't i supposed to make the equation =0?
question... is it sin(2x) or is it sin^2(x) for the problem...
if its sin^2(x)...its a totally different process lol
oh it's sin^2(x)
lol... 7 sin^2(x) - 14 sin x + 2 = -5 move -5 to other side to get: 7 sin^2(x) - 14 sin x + 7 = 0 Divide everything by 7 to get sin^2(x) - 2sin x + 1 = 0 Correct?
yea i get it better now
The trick here is now to take a variable such as B and set that equal to sin(x) so B=sin(x) Thus the equation becomes B^2-2B+1=0
Do you know how to factor this?
The equation becomes: (B-1)*(B-1)=0
Have you learnt multiplicity yet?
so so i do (sinx - 1)(sinx -1)=0 ?
so what does sin(x)=?
but why is it -1 in the parenthesis?
sinx = 1
its (sin(x) - 1)(sin(x) -1)=0
since sinx=1 what does x=?
do i do sin^-1 (1)? and thats equal pi/2?
yep! except theres a catch...
The question asks to find "all solutions"
thus (pi/2)+2pi is also a solution, cause if you go all the way around the circle it still causes sin(x) to equal 1. Thus the real answer is: (pi/2)+2pi*k where k is any positive or negative integer constant
(yes you'd have to write that full answer down on your hw to receive full credit)
oh i understand, thank you for your help!