Help With Trigonometric Equations?

Question

Hi, I have 3 equations that I need help with. Thanks to everyone who helps.

1. 1-4cos^2x=0
2. 3cot^2x-1=0
3. √2 sinx-2sin^2x=0

Answer 1

Note: I'm going to assume an interval of 0 <= x < 2pi.

1)

1 - 4cos^2(x) = 0

Notice that this is a difference of squares, so factor it as such and solve.

[1 - 2cos(x)] [1 + 2cos(x)] = 0

Equate each factor to 0.

1 - 2cos(x) = 0
1 + 2cos(x) = 0

cos(x) = 1/2
cos(x) = -1/2

cos(x) is equal to (1/2) at the points pi/3, 5pi/3.
cos(x) is equal to (-1/2) at the points 2pi/3, 4pi/3

Therefore, the solutions are x = {pi/3, 5pi/3, 2pi/3, 4pi/3}.

2)

3cot^2(x) - 1 = 0

Move the 1 to the right hand side.

3cot^2(t) = 1

Divide both sides by 3.

cot^2(t) = 1/3

Take the square root of both sides; it is important to include "plus or minus" when doing so.

cot(t) = sqrt(1/3)
cot(t) = 1/sqrt(3)

We can invert both sides, nothing that cotangent is the reciprocal of tangent.

tan(t) = sqrt(3)

tan(t) is equal to sqrt(3) at the points x = {pi/3, 4pi/3}

3)

√2 sin(x) - 2sin^2(x) = 0

First, factor out sin(x).

sin(x) [ √2 - 2sin(x) ] = 0

Equate each factor to 0.

sin(x) = 0
√2 - 2sin(x) = 0

sin(x) = 0
-2sin(x) = √2

sin(x) = 0
sin(x) = -√2/2

sin(x) is equal to 0 at the points {0, pi}.
sin(x) is equal to -√2/2 at the points 3pi/4 and 7pi/4.

Therefore, x = {0, pi, 3pi/4, 7pi/4}

Answer 2

1.)
1 - 4cos(x)^2 = 0
-4cos(x)^2 = -1
cos(x)^2 = (1/4)
cos(x) = (±1/2)

now depending if you want the answer to both the negative and positive, then

x = 60° or 300±°
x = 120° or 240°

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2.)
3cot(x)^2 - 1 = 0
3cot(x)^2 = 1
cot(x)^2 = (1/3)
cot(x) = 1/sqrt(3)
tan(x) = ±sqrt(3)

same reason above

x = 60° or 240°
x = 300° or 120°

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3.) If by this you mean
sqrt(2)sin(x) - 2sin(x)^2 = 0
(sin(x))(sqrt(2) - 2sin(x)) = 0

sin(x) = 0
x = 0°

sqrt(2) - 2sin(x) = 0
-2sin(x) = -sqrt(2)
sin(x) = (sqrt(2)/2)

Same reason above

x = 0°
x = °45°, or 135°
or
x = 225° or 315°

Answer 3

(1) 4cos^2x = 1 ==> cos^2x = 1/4 ==> cos x = +/-1/2 ==> x = +/- (pi/3) or +/- (2pi/3)

(2) cot^2x = 1/3 ==> cot x = +/- 1/sqrt(3) ==> x = +/- (pi/3) or +/- (2pi/3)

(3) sin x = sqrt(2)*sin^2x ==> sin x = 0 or sin x = 1/sqrt(2) = sqrt(2)/2 ==> x = 0 or pi or (pi/4) or (3pi/4)

Answer 4

1. 4(cosx)^2=1
=> (cosx)^2=(1/2)^2
=>cosx = 1/2, -1/2
=>x=npi +- pi/3 (Gnrl. Solution) = pi/3,2pi/3,5pi/3,7pi/3

2.Similarly:
cotx=1/sqrt3, -1/sqrt3
x=npi +- pi/3 (Gnrl. Solution) = pi/3,2pi/3,5pi/3,7pi/3

3.sinx(sqrt2 - 2sinx)=0
=> sinx=0,1/sqrt2
x=n pi, np + (-1)^n(pi/4) (GS)= 0, pi, pi/4, 3pi/4

Hope this helps!

Answer 5

1
(1-2cosx)(1+2cosx)=0
cosx=+/- 1/2

2
cotx=+/- 1/√3

3
sinx=1/√2, 0