Determine exact solutions for each equation with the interval xe[0, 2pi]

Question

4cos²x+2cosx=0

9csc²x-12=0

Answer 1

Need to know reference angles, the standard 30-60-09 and 45-45-90 triangles, and SOH-CAH-TOA.

Tip: always try to factor

0= 4cos²(x)+2cos(x) = 2cos(x)(2cos(x)+1) ⇒ cos(x)=0 and cos(x) = -1/2

For the first equation, this holds true for x=π/2, 3π/2.

For the second equation, consider the reference angle θ of x which exists in Q1 and has value cosθ=1/2 (all values are positive in Q1). By considering one of the special triangles, we find that θ=π/3. But this is only the reference angle, not x. Since cos(x)<0, this means x lives in either Q2 or Q3. Knowing this, we can find the angles that live in these quadrants and that have reference angle π/3.

In Q2, π-x=π/3 ⇒ x = 2π/3 and in Q3, x-π=π/3 ⇒ x = 4π/3. So the value set of solutions in [0,2π] are {π/2, 3π/2, 2π/3, 4π/3}.

A similar process holds for the second equation. Hint: factor!

Answer 2

4cos²x+2cosx=0

To solve this equation factor out the common factor 2cosx.

This gives: 2cosx ( 2cosx + 1) = 0

Next set each factor equal to zero: 2cosx = 0 and 2cosx + 1 = 0.

Solve for x: 2cosx = 0 2cosx + 1 = 0

cos x = 0 2cosx = -1

cosx = -1/2

In the interval from [0, 2π] where is cos x = 0? It is found at π/2 and 3π/2.

In the interval from [0, 2π], where is cos x = -1/2? It is found at 2π/3 and 5π/3

9csc²x-12=0

Add 12 to both sides: 9csc²x = 12

Divide by 9 csc²x = 12/9

Write in lowest terms: csc²x = 4/3

Take the square root of both sides: csc x = ±2/√3

If csc x = ±2/√3 , then sin x = ±√3/2, since they are reciprocal functions.

Where is sin x = ±√3/2 in the interval [0, 2π]?

sin x = √3/2 at π/3 and 2π/3

sin x = -√3/2 at 4π/3 and at 5π/3

Answer 3

Determine exact solutions for each equation with the interval [0, 2π]. a) 4 cos² x + 2 cos x = 0. b) 9 csc² x – 12 = 0

Detailed Solution:

a) 4 cos² x + 2 cos x = 0 [0, 2π]

4 cos² x + 2 cos x = 0 ==> 2 cos x (2 cos x + 1) = 0

2 cos x = 0 ==> cos x = 0 ==> x = cos^-1 0 ==> x = π/2 and x = 3π/2

2 cos x + 1 = 0 ==> 2 cos x = –1 ==> cos x = –1/2 ==> x = cos^-1 (–1/2) (Q2 & Q3)

Quadrant 2: x = cos^-1 (–1/2) = π – π/3 = 2π/3

Quadrant 3: x = cos^-1 (–1/2) = π + π/3 = 4π/3

Therefore: x = {π/2, 2π/3, 4π/3, 3π/2}

b) 9 csc² x – 12 = 0 [0, 2π]

9 csc² x – 12 = 0 ==> 9 csc² x = 12 ==> csc² x = 12/9 ==> csc² x = 4/3 ==> sin² x = 3/4 ==>

sin x = ±(√3)/(2) ==> sin x = (√3)/(2) (Q1 & Q2) ==> sin x = – (√3)/(2) (Q3 & Q4)

Quadrant 1: sin x = (√3)/(2) ==> x = sin^-1 (√3)/(2) ==> x = π/3

Quadrant 2: sin x = (√3)/(2) ==> x = sin^-1 (√3)/(2) ==> x = π – π/3 ==> x = 2π/3

Quadrant 3: sin x = –(√3)/(2) ==> x = sin^-1 –(√3)/(2) ==> x = π + π/3 ==> x = 4π/3

Quadrant 4: sin x = –(√3)/(2) ==> x = sin^-1 –(√3)/(2) ==> x = 2π – π/3 ==> x = 5π/3

Therefore: x = {π/3, 2π/3, 4π/3, 5π/3}