Question

cos 3 theta + cos theta - 2 cos theta=0

Answer 1

Correct question :

Solve for ∅ in the given equation :-

cos3∅ + cos∅ – 2cos2∅ = 0

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Answer:

♦ Principal solution :-

∅ = 0 , π/4 , 3π/4 , 2π

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♦ General solution :-

∅ = nπ ± π/4 or 2mπ where m , n € Z

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Note:

★ sin(A + B) = sinA•cosB + cosA•sinB

★ sin(A – B) = sinA•cosB – cosA•sinB

★ cos(A + B) = cosA•cosB – sinA•sinB

★ cos(A – B) = cosA•cosB + sinA•sinB

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★ sin(A + B) + sin(A – B) = 2sinA•cosB

★ sin(A + B) – sin(A – B) = 2cosA•sinB

★ cos(A + B) + cos(A – B) = 2cosA•cosB

★ cos(A + B) – cos(A – B) = - 2sinA•sinB

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★ $sinC + sinD = 2.sin( \frac{C+ D}{2} ).cos(\frac{C-D}{2})$

★ $sinC - sinD = 2.cos( \frac{C+ D}{2} ).sin(\frac{C-D}{2})$

★ $scosC + cosD = 2.cos( \frac{C+ D}{2} ).cos(\frac{C-D}{2})$

★ $cosC + cosD = - 2.sin( \frac{C+ D}{2} ).sin(\frac{C-D}{2})$

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★ If sin∅ = sinα , then ;

∅ = nπ + (-1)ⁿα , n € Z

★ If cos∅ = cosα , then ;

∅ = 2nπ ± α , n € Z

★ If tan∅ = tanα , then ;

∅ = nπ + α , n € Z

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Solution:

• Given: cos3∅ + cos∅ – 2cos2∅ = 0
• To find : ∅ = ?

We have ;

=> cos3∅ + cos∅ – 2cos2∅ = 0

=> [ cos3∅ + cos∅ ] – 2cos2∅ = 0

=> 2cos{ (3∅ + ∅)/2 }•cos{ (3∅ - ∅)/2 }

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$ – 2cos2∅ = 0

=> 2cos(4∅/2)•cos(2∅/2) – 2cos2∅ = 0

=> 2cos2∅•cos∅ – 2cos2∅ = 0

=> 2cos2∅•( cos∅ – 1 ) = 0

=> cos2∅•( cos∅ – 1 ) = 0

Here,

Two cases arises :-

★ Case(1) : cos2∅ = 0

• Principal solution :-

=> cos2∅ = 0

=> cos2∅ = cosπ/2 or cos(2π - π/2)

=> cos2∅ = cosπ/2 or cos3π/2

=> 2∅ = π/2 or 3π/2

=> ∅ = π/4 or 3π/4

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• General solution :-

=> cos2∅ = 0

=> cos2∅ = cosπ/2

=> 2∅ = 2nπ ± π/2 , n € Z

=> ∅ = nπ ± π/4 , n € Z

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★ Case(2) : cos∅ – 1 = 0

• Principal solution :-

=> cos∅ – 1 = 0

=> cos∅ = 1

=> cos∅ = cos0 or cos(2π - 0)

=> cos∅ = cos0 or cos2π

=> ∅ = 0 or 2π

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• General solution :-

=> cos∅ – 1 = 0

=> cos∅ = 1

=> cos∅ = cos0

=> ∅ = 2mπ ± 0 , m € Z

=> ∅ = 2mπ , m € Z

Answer 2

Answer:

1. cos3theta +cos theta-2cos theta =0