# 1. 12cosx-5=4cosx 2. 9sin^2x-9sinx= -2 3. 3tanx=Cotx Find the solution for these trigonometric equations ?

### 2 Answers

- la consoleLv 71 month ago
12.cos(x) - 5 = 4.cos(x)

12.cos(x) - 4.cos(x) = 5

8.cos(x) = 5

cos(x) = 5/8

x = ± arccos(5/8)

x ≈ ± 0.895664 + 2kπ → where k is an integer

9.sin²(x) - 9.sin(x) = - 2

sin²(x) - sin(x) = - 2/9

sin²(x) - sin(x) + (1/2)² = - (2/9) + (1/2)²

sin²(x) - sin(x) + (1/2)² = 1/36

[sin(x) - (1/2)]² = 1/36

sin(x) - (1/2) = ± 1/6

sin(x) = (1/2) ± (1/6)

sin(x) = (3 ± 1)/6

First case: sin(x) = (3 + 1)/6

sin(x) = 4/6

sin(x) = 2/3

x = arcsin(2/3)

x₁ ≈ 0.729727 + 2kπ → where k is an integer

x₂ ≈ (π - 0.729727) + 2kπ → where k is an integer

Second case: sin(x) = (3 - 1)/6

sin(x) = 2/6

sin(x) = 1/3

x = arcsin(1/3)

x₁ ≈ 0.339836 + 2kπ → where k is an integer

x₂ ≈ (π - 0.339836) + 2kπ → where k is an integer

3.tan(x) = cot(x)

3.[sin(x)/cos(x)] = cos(x)/sin(x)

3.sin²(x) = cos²(x) → recal: cos²(x) + sin²(x) = 1 → sin²(x) = 1 - cos²(x)

3.[1 - cos²(x)] = cos²(x)

3 - 3.cos²(x) = cos²(x)

4.cos²(x) = 3

cos²(x) = 3/4

cos(x) = ± (√3)/2

First case: cos(x) = (√3)/2

x₁ = (π/3) + 2kπ → where k is an integer

x₂ = (5π/3) + 2kπ → where k is an integer

Second case: cos(x) = - (√3)/2

x₃ = (2π/3) + 2kπ → where k is an integer

x₄ = (4π/3) + 2kπ → where k is an integer

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- PhilipLv 61 month ago
Put (s,c) = [sinx,cosx].

1. 12c-5 = 4c, ie., 8c=5, c = (5/8), x = arccos(5/8) = 51.31781255°.

2. 9s^2-9s = -2, ie., s^2 -s = -(2/9). Then s^2 -s +(1/4) = (1/4)-(2/9) =1/36 = (1/6)^2

and [s-(1/2)]^2 = (1/6)^2, ie., s = (1/2)(+/-)(1/6) = (1/6)[3(+/-)1] = (1/3)..(i) or (2/3)..(ii).

For (i) holding, x = arcsin(1/3) = 19.47122062° or [180° -arcsin(1/3)] = 160.5287794°.

For (ii) holding, x = arcsin(2/3)= 41.81031490° or [180° -arcsin((2/3)] =138.1896851°.

3. 3(s/c) = (c/s), ie., 3 = c^2/s^2, ie., 3s^2 = c^2 = 1-s^2, ie., s = (+/-)(1/2).

For s = (1/2), x = 30° or 150°.

For s = (-1/2), x = (180+30)° or (360-30)° = 210° or 330°.

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