Solve 7 cos(2 theta) = 7 sin^2(theta) + 3 for all solutions 0 less than or equal to theta less than 2pi.

Solve 7 sin(2 w) - 4 cos(w) =0 for all solutions 0 le w lt 2pi

le= less than or equal to

lt = less than

I keep getting numbers that are not on the unit circle and then when I try to plug them into my calculator I'm getting really big numbers. Please help. I've tried looking in my book but there are no problems like these.

7 cos(2 theta) = 7 sin^2(theta) + 3

7 cos^2 Ø - 7sin^2 Ø = 3
cos^2Ø - sin^2 Ø = 3/7
cos 2Ø = 3/7

so 2Ø is in I or in IV
set your calculator to radians
2Ø = 1.1279 or 2Ø = 2π-1.1278 = 5.1553
Ø = .5639 or Ø = 2.578
Now the period of cos 2Ø = π , so by adding multiples of π to any answer will yield a new answer

so Ø = .5639+π = 3.7055
Ø = 5.1553+π = 5.71924


7 sin 2w - 4cos w = 0
7(2sinwcosw) - 4cosw = 0
cosw(14sinw - 4 ) = 0
cosw = 0 or sinw = 2/7

if cosw = 0, then w = π/2 , 3π/2
if cosw = 2/7 , then w = 1.281 or 2π - 1.281 = 5.002

Thank you for your help! My numbers were way off haha.

To solve the equation 7*cos(2*theta) = 7*sin^2(theta) + 3, we can first simplify the equation and then solve for theta.

Step 1: Simplify the equation
Using the trigonometric identity sin^2(theta) = 1 - cos^2(theta), we can rewrite the equation as:

7*cos(2*theta) = 7*(1 - cos^2(theta)) + 3

Step 2: Distribute and rearrange terms
Expanding the equation:

7*cos(2*theta) = 7 - 7*cos^2(theta) + 3

7*cos^2(theta) + 7*cos(2*theta) = 10

Step 3: Apply the double-angle identity
Using the double-angle identity cos(2*theta) = 2*cos^2(theta) - 1, we can substitute it into the equation:

7*cos^2(theta) + 7*(2*cos^2(theta) - 1) = 10

Step 4: Distribute and rearrange terms
Expanding the equation:

7*cos^2(theta) + 14*cos^2(theta) - 7 = 10

21*cos^2(theta) = 17

Step 5: Solve for cos^2(theta)
Divide both sides of the equation by 21:

cos^2(theta) = 17/21

Step 6: Take the square root
Taking the square root of both sides of the equation:

cos(theta) = ±√(17/21)

Step 7: Solve for theta
Since we are interested in all solutions for 0 ≤ theta < 2π, we can apply the inverse cosine function to find theta:

theta = arccos(√(17/21)) and theta = arccos(-√(17/21))

Using a calculator, we can find the numerical values of these angles. Remember to take into account the given range of 0 ≤ theta < 2π.

Similarly, for the equation 7*sin(2*w) - 4*cos(w) = 0, you can follow the same steps to solve for w:

Step 1: Simplify the equation
Step 2: Distribute and rearrange terms
Step 3: Apply the double-angle identity
Step 4: Distribute and rearrange terms
Step 5: Solve for sin^2(w)
Step 6: Take the square root
Step 7: Solve for w

By following these steps, you can find all the solutions for 0 ≤ w < 2π.