Science & Mathematics » Mathematics » 2(sin(theta))^2=3(1-cos(theta)...

2(sin(theta))^2=3(1-cos(theta)...

How do I find the answer to this? Thanks

• Pythagorean identity:

sin^2 (theta) + cos^2(theta) = 1

sin^2(theta) = 1 - cos^2(theta)

2(sin(theta))^2=3(1-cos(theta))

2(1-cos^2(theta)) = 3(1-cos(theta))

Divide by (1-cos(theta)) each side

2 (1-cos(theta)) = 3

1 - cos(theta) = 3/2

1 = 3/2 + cos(theta)

1 - 3/2 = cos(theta)

arcos(-1/2) = theta

Arcos is sometime seen as ''cos^-1'' on calculators.
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• (sin(theta))^2 can be written as 1 - (cos(theta))^2
then assume cos(theta)=x
2(1-x^2) = 3(1-x)
2x^2-3x+1 = 0
x=1/2 or x=1
therefore theta can be 0 , 60 or 300 degrees
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• 2(sin^2(x)) = 2(1 - cos^2(x))
so
2sin^2(x) = 3(1 - cos(x))
==> 2 - 2cos^2(x) - 3 + 3cos(x) = 0
==> 2cos^2(x) - 3cos(x) + 1 = 0
==> (2cos(x) -1)(cos(x) - 1) = 0
==> cos(x) = 1/2 or cos(x) = 1
So, x = +/- pi/3 or x = 0 (you can add the other solutions with multiples of 2pi)
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• Your qn floored me at first - are attempting squaring all sides (if you happen to see a sq root are trying, this will simplify matters) instantly you get: ( if c = cos theta, s = sin theta) c^2 + 2cs + s^2 = 2c^2 or c^2 - s^2 = 2cs now consider half angles.... The lhs (left hand facet) is the same as cos(2.Theta) the rhs is the same as sin(2.Theta) Now sin(2.Theta) = cos(2.Theta) at forty five levels, i.E 2.Theta=45 = pi/4 (radians) so answer is theta = pi/eight
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• Im going to use "x" instead of theta. Also I write sin^2x for (sinx)^2.

2sin^2x=3(1-cosx) [we make the substitution sin^2x=1-cos^2x] *Also we can see here 0 is already a solution. Because 2sin^2(0)=0 and 3(1-cos(0))=0

2(1-cos^2x)=3(1-cosx) [ simplify and get a quadratic equation]

2-2cos^2x=3-3cosx

2=2cos^2x-3cosx+3

0=2cos^2x-3cosx+1. [ Now let cosx= z ]

z=[-b+-sqr(b^2-4ac)]/2a [where, a=2,b=-3,and c=1.]

z=[3+-sqr(9-4(2)(1)]/4

we get two solutions as usual. z1= 1 and z2=1/2.

that is we have cosx=1 and cosx=1/2.

cosx=1 when x= 0+2pik, where k={...-3,-2,-1,0,1,2,3,...} (that is k is any integer)

cosx=1/2 when x=+-pi/3+2pik, and once again k is any integer. (thats plus or minus pi/3)

Lets check our solutions!

2sin^2(pi/3)=3(1-cos(pi/3)) =3/2 TRUE.

2sin^2(-pi/3)=3(1-cos(-pi/3))=3/2 TRUE.

2sin^2(0)=3(1-cos(0))=0 TRUE.

Note: The general solutions are x1=0+2pik x2=pi/3+2pik , and x3=-pi/3+2pik
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