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Question

If α,β are the solutions of sinx=12 in [0,2π] and α,γ are the solutions of cosx=32 in [0,2π], then

A
αβ=π
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B
βγ=π
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C
αγ=π
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D
α+β=3π
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Solution

The correct options are
B βγ=π
D α+β=3π
Solutions of sinx=12 in [0,2π] are
x=π+π6 and 2ππ6
x=7π6 and 11π6 ...(1)

Solutions of cosx=32 in [0,2π] are
x=π±π6=5π6,7π6 ...(2)

From (1) and (2),
α=7π6, β=11π6, γ=5π6
α+β=3π
αβ=2π3
βγ=π

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