Check answers please ( take home quiz)

Question

1-4 solve each for 0 < or equal to x < 2pi (radians)

 

1.  8 cos ^2 (x) - 1 = 5 

 

A: 30, 150, 210, 330

x=     pi/6, 5/6 pi, 7/6 pi, 11/6 pi

 

2. -4 csc (x) = 8 

A: 210, 330

  x=   7/6 pi, 11/6 pi

 

3. 2 cos^2 (2x)= 2 cos (2x) (not really sure...)

A: x= pi/4, pi/2 , 3/4 pi 

 

4. sin (2x)= 2 cos (x)

A: 90, 270 

x= pi/2 , 3/2 pi 

 

 

5-6 solve each for 0 < or wqual to x < 360 

 

5. 8 tan (x-15)=8

A: x= 60

 

6. 2 cos ^2 (x) + cos (x)= 0 

A: x= 90, 120,240,270

 

 

7-9 solve each for all solutions of x in radians. 

 

7. 2 sec (x) -3=1

A: pi/3, 5/3 pi  (+ 2npi)

 

 

8. cot(2x)= radical 3 

A: pi /12, 7/12 pi (+2npi)

 

 

9. tan (X) - 2 sin(x)tan(x)= 0

A: 0, pi/6, 5/6 pi, pi (+2npi)

 

Answer 1

Question 1

 

8cos²(x) - 1 = 5 

 

Add 1 on both sides of equation.

 

 

8cos²(x) = 6

 

 

Divide both sides of equation by 8.  Then reduce the right side of equation.

 

cos²(x) = 3/4

 

 

Square-root both sides of equation.

 

cos(x) = ± √(3) / 2

 

cos(x) = √(3) / 2          and      cos(x) = -√(3) / 2

 

 

Cosine is only positive in the 1st and 4th quadrants.

 

x = 30                         and      x = 180 - 30 = 150

x = 330                       and      x = 180 + 30 = 210

 

 

We have four solutions of x.

 

 

Question 2

 

-4csc(x) = 8

 

 

Divide both sides of equation by -4.

 

 

csc(x) = -2

 

csc(x) is the same as 1 / sin(x).

 

 

1 / sin(x) = 2

 

1 = 2sin(x)

 

sin(x) = 1 / 2

 

x = 30

 

sine is only positive in the 1st and 2nd quadrant.

 

x = 180 - 30

x = 150

 

The solutions are

 

x = 30      and      x = 150

 

 

Question 3     (this is the one that troubles you the most I believe)

 

2cos²(2x) = 2cos(2x)

 

 

Divide both sides of equation by 2.

 

cos²(2x) = cos(2x)

 

 

cos(2x) is a double angle that is equivalent to cos²(x) - sin²(x).  

 

Also, we have the identity   sin²(x) + cos²(x) = 1.  We can utilize this.  Let's get everything in terms of cosine.

 

 

[cos²(x) - (1 - cos²(x))]² = cos²(x) - (1 - cos²(x))

 

(2cos²(x) - 1)² = 2cos²(x) - 1

 

 

Make the right side to equal zero.

 

(2cos²(x) - 1)² - 2cos²(x) + 1 = 0

 

(2cos²(x) - 1)(2cos²(x) - 1) - 2cos²(x) + 1 = 0

 

4cos⁴(x) - 4cos²(x) + 1 - 2cos²(x) + 1 = 0

 

4cos⁴(x) - 6cos²(x) + 2 = 0

 

2(2cos⁴(x) - 3cos²(x) + 1) = 0

 

2cos⁴(x) - 3cos²(x) + 1 = 0

 

 

Let   q² = cos⁴(x)  and  q = cos²(x)  to make solving this quadratic equation easier.

 

    

2q² - 3q + 1 = 0

(2q - 1)(q - 1) = 0

 

q = 1/2     and    q = 1

 

 

Substitute q into the cosine function to solve for x.

 

cos²(x) = 1/2           and          cos²(x) = 1

 

cos(x) = ± √(2) / 2        and       cos(x) = ± 1

cos(x) = √(2) / 2            and           x = 0 

cos(x) = -√(2) / 2                           x = 180

                                                     x = 360

x = 45

 

For positive cosine:

 

x = 360 - 45 = 315

 

 

For negative cosine:

 

x = 180 - 45 = 135

x = 180+ 45 = 225

 

 

There are seven solutions.

 

x = 0

x = 45

x = 135

x = 180

x = 225

x = 315

x = 360

  

 

Now that you know how to tackle these problems, try the others on you own. Use this identity for the next equations:

 

tan(x + y) = (tan(x) + tan(y)) / (1 - tan(x)tan(y))

tan(x - y) = (tan(x) - tan(y)) / (1 + tan(x)tan(y))