Hayya please help with mine its right above
4sin²(x) = 5 - 4cos(x)
4{¹/₂[1 - cos(2x)]} = 5 - 4cos(x)
4{¹/₂[1] - ¹/₂[cos(2x)]} = 5 - 4cos(x)
4[¹/₂ - ¹/₂cos(2x)] = 5 - 4cos(x)
4[¹/₂] - 4[¹/₂cos(2x)] = 5 - 4cos(x)
2 - 2cos(2x) = 5 - 4cos(x)
- 2 - 2
-2cos(2x) = 3 - 4cos(x)
-2[2cos²(x) - 1] = 3 - 4cos(x)
-4cos²(x) + 2 = 3 - 4cos(x)
- 2 - 2
-4cos²(x) = 1 - 4cos(x)
-4cos²(x) + 4cos(x) - 1 = 0
4cos²(x) - 4cos(x) + 1 = 0
[2cos(x) - 1]² = 0
2cos(x) - 1 = 0
+ 1 + 1
2cos(x) = 1
2 2
cos(x) = ¹/₂
cos⁻¹[cos(x)] = cos⁻¹(¹/₂)
x = 60, 300
x = π/3, 5π/3
[0, 2π) = 0 ≤ x < 2π
[0, 2π) = 0 ≤ π/3 ≤ 2π or 0 ≤ 5pi/3 < 2π
13. 0.0000000228
14. 1 over 15
15. one-third
16. point three repeating
17. 2two and five eighths
18. -0.59
19. 2.25
20.
21. -0.25
22. 0.6
23. 3
24. 4 and 4 over 9
D: because there are 2 4/4 which is 2 plus 14/4 bc that’s the rest added up so all together that is 22/4 which reduced by half is 11/2
Answer:
he answer is in the image
Step-by-step explanation:
