ahh! need help for this question urgently! (: thanks!?

Question

Solve the following equations for
0degrees
1) sin 2Q+ sin 4Q=0

2) cos 4Q- cos 6Q=0

3) 2 sinQcosQ+ cos^2 Q-1=0

thank you soooo much!

Answer 1

Use double angle formulas to break them down to angles of Q, not multiples of Q.

1) sin2Q + sin4Q=0
2sinQcosQ + 2sin2Qcos2Q=0
2sinQcosQ + 4sinQcosQ(2(cosQ)^2-1)=0
Factor
[2sinQcosQ](1 + 2(2(cosQ)^2-1)=0
[2sinQcosQ](4(cosQ)^2-1)=0
Set each term equal to zero
sinQ=0
cosQ=0
4(cosQ)^2 - 1=0

Q=0, 180
Q=90, 270
Q=arccos(1/2)=60, 120

I stared at the second one for awhile and couldn't come up with anything simple to solve it. I'm sure there is some trick I'm just not seeing right now. Good luck on that one.

3) The trick here is to substitute for the last two terms:
(cosQ)^2 - 1 = -(sinQ)^2
2sinQcosQ-(sinQ)^2 = 0
sinQ[2cosQ-sinQ]=0
sinQ=0
2cosQ-sinQ=0
Q=0, 180 for the first part.
2cosQ=sinQ
2=sinQ/cosQ
2=tanQ
Q=arctan(2)=63.43 and 243.43

Viola! Good luck with number 2...

Answer 2

Sin(2x) = 2Sin(x)Cos(x)
Cos(2x) = [Cos(x)]^2 - [Sin(x)]^2
[Sin(x)]^2 + [Cos(x)]^2 = 1

Using that,
sin 2Q+ sin 4Q=0
sin 2Q + 2 sin 2Q cos 2Q = 0
sin 2Q (1 + 2 cos 2Q) = 0
therefore:
either Sin 2Q = 0 (2Q = 0 or 180 or 360),

or (1 + 2 cos 2Q) = 0
2 cos 2Q = -1
cos 2Q = -1/2
2Q = 120 degrees or 240 degrees
Q = 60 degrees or 120 degrees

Q = 0, 60, 90, 120, 180...

2) bring everything to x = 2Q

3)1 - cos^2 = sin
2 sinQcosQ - (1- cos^2)
2 sinQcosQ - sin^2 = 0
multiply everything by -1 (both sides) and rearrange:
sin^2 + (2cos)sin + 0 = 0
sin = [-2cos +/-SQRT(4cos^2)]/2
sinQ = 0 or 2cosQ
sinQ = 2cosQ, divide both sides by cosQ
sin/cos = 2
tanQ = 2
Q = 63.435 degrees

Answer 3

1. sin 2Q+ sin 4Q=0
sin 4Q = sin(2(2Q)) = 2 sin2Q cos2Q
sin 2Q = -2 sin2Q cos 2Q
so cos 2Q = -1/2 and 2Q = 120 or 240 degrees
Q = 60 or 120 degrees

Also, since there is a sin 2Q on both sides, sin 2Q = 0
this gives Q = 0 Q = 90 Q = 180 degrees

So Q = 0, 60, 90, 120, 180 degrees

2. cos 4Q- cos 6Q=0

cos 6Q = cos(4Q + 2Q) = cos4Q cos 2Q - sin 4Q sin 2Q
Put this in original equation and divide by cos 4Q
1 - cos 2Q + sin 2Q tan 4Q = 0 [EQ - 1]

cos 4Q = cos(6Q - 2Q) = cos 6Q cos 2Q + sin 6Q sin 2Q
Put this in original equation and divide by cos 6Q
cos 2Q + tan 6Q sin 2Q - 1 = 0 or
1 - cos 2Q = tan 6Q sin 2Q [EQ - 2]

Combine [EQ - 1] and [EQ - 2]
tan 6Q sin 2Q + tan 4Q sin 2Q = 0
sin 2Q (tan 6Q + tan 4Q) = 0
so
sin 2Q = 0 gives one set of answers Q = 0, 90, 180, 270

Must also solve:
tan 6Q + tan 4Q = 0
tan 6Q = tan(4Q + 2Q) = (tan 4Q + tan 2Q)/( 1 - tan 4Q tan 2Q)
tan 4Q + tan 2Q + tan4Q (1 - tan 4Q tan 2Q) = 0
2 tan 4Q + tan 2Q - tan^2 4Q tan 2Q = 0
tan 4Q (2 - tan 4Q tan 2Q) + tan 2Q = 0

tan 4Q = tan(2(2Q)) = 2 tan 2Q / (1 - tan^2 2Q)

2 tan 2Q (2 - 4 tan^2 2Q) + tan 2Q( 1 - 2tan^2 2Q + tan^4 2Q) = 0
5 tan 2Q - 10 tan^3 2Q + tan^5 2Q = 0
5 - 10 tan^2 2Q + tan^4 2Q = 0
tan^2 2Q = 5 + SQRT(20) using (-b +/- SQRT(b^2 - 4ac)) / 2a
tan 2Q = +/- SQRT(5 + SQRT(20)) = +/- SQRT(9.472) = +/-3.0777
Q = 36, 144, 216, 324

Answers are 0, 36, 90, 144, 180 , 216, 270, 324

3. 2 sinQcosQ+ cos^2 Q-1=0
2 sinQcosQ = 1 - cos^2 Q = sin^2 Q
2 cosQ = sinQ and 2 = tanQ
Q = 63.435 degrees or 243.435 degrees