Question

4cos Square theta -4 sin theta -1 = 0 where theta is positive acute angle

Answer 1

Solution

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As sin

2

θ+cos

2

θ=1

⇒sin

2

θ=1−cos

2

θ

⇒4sin

2

θ−8cosθ+1=0

⇒4(1−cos

2

θ)−8cosθ+1=0

⇒−4cos

2

θ−8cosθ+5=0

⇒−4cos

2

θ−10cosθ+2cosθ+5=0

⇒−2cosθ(2cosθ+5)+(2cosθ+5)

⇒(−2cosθ+1)(2cosθ+5)=0

⇒cosθ=

2

1

or cosθ=

2

−5

⇒θ=60

o

∣cosθ∣≤1.

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