Find the exact solutions of the equation that are in the interval [0, 2*Pi):?

Question

sin(2t) + sin(t) = 0

cos(u) + cos(2u) = 0

tan(2u) = tan(u)

Answer 1

sin(2t) = 2sin(t)cos(t), so

sin(2t) + sin(t) = 0 => 2sin(t)cos(t) + sin(t) = 0
This is true if sin(t) = 0.
If sin(t) != 0 then divide by sin(t) to get
2cos(t) + 1 = 0
cos(t) = -1/2
So the solutions are t = 0, t = pi, t = 5pi/6, t = 7pi/6

cos(2u) = cos(u)^2 - sin(u)^2 = 2cos(u)^2 - 1, so
cos(u) + cos(2u) = 0 => cos(u) + 2cos(u)^2 - 1 = 0
let x = cos(u) then this is 2x^2 + x - 1 = 0
This factors to (2x - 1)(x + 1) = 0

so cos(u) = 1/2 or cos(u) = -1
u = pi/3 or 5pi/3 or pi

tan(2u) = sin(2u)/cos(2u) = 2sin(u)cos(u)/(cos(u)^2 - sin(u)^2) so
sin(u)/cos(u) = 2sin(u)cos(u)/(cos(u)^2 - sin(u)^2)
Cross multiply:
cos(u)^2sin(u)-sin(u)^3 = 2sin(u)cos(u)^2

If sin(u) = 0 this is true, so 0 and pi are 2 answers.
Otherwise divide by sin(u):
cos(u)^2 - sin(u)^2 = 2cos(u)^2
-sin(u)^2 = cos(u)^2 has no solutions (negative = positive), so the above are the only answers.

I hope there are no careless errors here, you'd better check to be sure.

Answer 2

a) Use the double-angle formula sin(2t) = 2sin(t)cos(t)
sin(2t) + sin(t) = 2sin(t)cos(t) + sin(t)
= sin(t)*[2cos(t) + 1] = 0
2 possibilities:
1) sin(t) = 0 ──> t = arcsin(0) = {0, π}
and
2) 2cos(t) + 1 = 0 ──> cos(t) = -1/2
──> t = arccos(-1/2) = {2π/3, 4π/3}
So your solutions are: t = {0, 2π/3, π, 4π/3}

b) Use the double-angle formula cos(2u) = cos²(u) - sin²(u)
And the pythagorean identity sin²(u) = 1 - cos²(u)
cos(2u) + cos(u) = [2cos²(u) -1] + cos(u)
= [2cos(u) - 1]*[cos(u) + 1] = 0
2 possibilities:
1) 2cos(u) - 1 = 0 ──> u = arccos(1/2) = {π/3, 5π/3}
2) cos(u) + 1 = 0 ──> u = arccos(-1) = π
So your solutions are: u = {π/3, π, 5π/3}

c) Use the double-angle formula tan(2u) = 2tan(u)/(1-tan²(u))
tan(2u)-tan(u) = 2tan(u)/(1-tan²(u)) - tan(u)
= tan(u)*[2/(1-tan²(u)) - 1] = tan(u)*[(2-1+tan²(u)) / (1-tan²(u))]
= tan(u)*[(1+tan²(u)) / (1-tan²(u))]
Now use the double-angle identity cos(2u) = (1+tan²(u)) / (1-tan²(u))
──> tan(2u) - tan(u) = tan(u)*[1 / cos(2u)] = 0
2 possibilities:
1) tan(u) = 0 ──> u = arctan(0) = {0, π}
2) [1 / cos(2u)] = 0
Since -1 ≤ cos(2u) ≤ +1, 1 / cos(2u) is never zero.
So your solutions are: u = {0, π}

Answer 3

sin(2t) + sin(t) = 0
2 sin(t)cos(t) + sin(t) = 0
sin(t)[2cos(t) + 1] = 0

so either sin(t)=0, in which case t=0 or Pi
or cos(t) = -1/2, in which case t=2Pi/3 or 4Pi/3
(using fact that cos(Pi/3)=1/2, and cos(Pi-t)=-cos(t) to get 2Pi/3. also used cos(t)=cos(-t) to get t=-2Pi/3, then cos(t)=cos(2Pi+t) to get 4Pi/3)

cos(u) + cos(2u) = 0
cos(u) + 2 cos^2(u) - 1 = 0
quadratic formula:
cos(u) = (-1 +/- sqrt(1 - 4*2*-1))/4
= (-1 +/- sqrt(9))/4 =(-1 +/- 3)/4 = -1 or 1/2
cos(u) = -1 means u=Pi
cos(u)=1/2 means u=Pi/3 or 5Pi/3

tan(2u) = tan(u)
sin(2u)/cos(2u) = sin(u)/cos(u)
sin(2u)cos(u) = sin(u)cos(2u)
2sin(u)cos^2(u) = sin(u)[2cos^2(u)-1]
2sin(u)cos^2(u) = 2sin(u)cos^2(u)-sin(u)
sin(u)=0, so u=0 or Pi

Answer 4

sin(2t) + sin(t) = 0
or sin(t)(cos(t)+1) = 0

So, t = 0,pi

cos(u) + cos(2u) = 0
or 2cos^2(u)+cos(u)-1 = 0
or (2cos(u)-1)(cos(u)+1) = 0
u = pi,pi/3,5pi/3

tan(2u) = tan(u)
2tan(u) = tan(u)-tan^3(u)
or

Answer 5

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