For your earlier post for this same question at 3:17 am, I said
"for #2 I too, like Bob, was puzzled
First of all is it
sin 2θ = cos θ or sin2 θ = cos θ ?
for the first one
2sinθcosθ - cosθ = 0
cosθ(1sinθ - 1) = 0
cosθ=0 or sinθ=1/2
so θ = 90º or θ = 30º or 150º for your domain, which is choice C "
how did you get your second line???
2)Find the solution of sin2 theta = cos theta if 0 -< theta < 180
A)30 degrees and 90 degrees
B)30 degrees and 150 degrees
C)30 degrees, 90 degrees, 150 degrees
D)0 degrees, 90 degrees, and 150 degrees
sin2 theta = cos theta
sin2 theta - 1 = 0
(sin theta + 1)(sin theta - 1)= 0
sin theta = -1 or sin theta = 1 which is B
I used the zero property
5 answers
line: cosθ(1sinθ - 1) = 0
should have said
cosθ(2sinθ - 1) = 0
should have said
cosθ(2sinθ - 1) = 0
Please use ^ in front of exponents. Otherwise, we cannot tell sin (2theta) from (sin theta)^2.
Your first step is wrong, since cos theta is not necessarly equal to 1. Also, if sin theta were 1 or -1, why would the answer be B?
I will assume you meant to write sin (2theta) and will let x be theta, to simplify typing.
2 sin x cos x = cos x
2 sin x = 1
sin x = 1/2
x = 30 or 150 degrees.
You seem to have treated sin2theta as sin^2 theta, in your next-to-last step, but if you do that, the answer is completely different
Your first step is wrong, since cos theta is not necessarly equal to 1. Also, if sin theta were 1 or -1, why would the answer be B?
I will assume you meant to write sin (2theta) and will let x be theta, to simplify typing.
2 sin x cos x = cos x
2 sin x = 1
sin x = 1/2
x = 30 or 150 degrees.
You seem to have treated sin2theta as sin^2 theta, in your next-to-last step, but if you do that, the answer is completely different
my question is sin 2è = cos è NOT sin2 è = cos è. I'm sorry, I didn't understand your question earlier.
In that case the solution I just gave you above will be the correct one.
Notice that all 3 answers work in your original equation.
Notice that all 3 answers work in your original equation.