cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = x
cos(x + x) = cos(x).cos(x) - sin(x).sin(x)
cos(2x) = cos²(x) - sin²(x)
cos(2x) = [1 - sin²(x)] - sin²(x)
cos(2x) = 1 - sin²(x) - sin²(x)
cos(2x) = 1 - 2.sin²(x) ← memorize this result
Your equation:
cos(2x) - 5.sin(x) - 3 = 0 → recall (1)
1 - 2.sin²(x) - 5.sin(x) - 3 = 0
- 2.sin²(x) - 5.sin(x) - 2 = 0
2.sin²(x) + 5.sin(x) + 2 = 0
2.sin²(x) + [4.sin(x) + sin(x)] + 2 = 0
2.sin²(x) + 4.sin(x) + sin(x) + 2 = 0
[2.sin²(x) + 4.sin(x)] + [sin(x) + 2] = 0
2.sin(x).[sin(x) + 2] + [sin(x) + 2] = 0
[sin(x) + 2].[2.sin(x) + 1] = 0
First case: [sin(x) + 2] = 0 → sin(x) + 2 = 0 → sin(x) = - 2 ← no posible
Second case: [2.sin(x) + 1] = 0 → 2.sin(x) + 1 = 0 → 2.sin(x) = - 1 → sin(x) = - 1/2
First possibility: x₁ = π + (π/6) → x₁ = 7π/6
Second possibility: x₂ = 2π - (π/6) → x₂ = 11π/6
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D.W. say: does "cos2x" mean cos²x, or cos(2x) ?
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stanschim say: cos(2x) - 5sin(x) - 3 = 0
(cos(x))^2 - (sin(x))^2 - 5sin(x) - 3 = 0
(1 - (sin(x))^2) - (sin(x))^2 - 5sin(x) - 3 = 0
-2(sin(x))^2 - 5sin(x)) -2 = 0
2(sin(x))^2 + 5sin(x) + 2 = 0
'
(2sin(x)) + 1)(sin(x) + 2) = 0
Solution occurs when 2sin(x) + 1 = 0. sin(x) + 2 can never equal 0 since that would require sin(x) = -2 (that is not possible).
2sin(x) + 1 = 0
2sin(x) = - 1
sin(x) = -1/2
Between 0 and 360 degrees, this will be satisfied at 210 degrees and 330 degrees. More generally, the solution could be expressed in degrees as (210 + 360(n-1)) degrees and (330 + 360(n-1)) degrees where n runs from 1,2,3...
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Jeff Aaron say: cos(2x) - 5*sin(x) - 3 = 0
1 - 2 sin^2(x) - 5*sin(x) - 3 = 0
Let u = sin(x), so we have:
1 - 2u^2 - 5u - 3 = 0
-2u^2 - 5u - 2 = 0