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Q.

If 0≤θ≤π and  81sin2⁡θ+81cos2⁡θ=30,then θ is

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a

30∘

b

60∘

c

120∘

d

150∘

answer is A.

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Detailed Solution

cos2⁡θ=1−sin2⁡θ. Let 81sin2⁡θ=z, we get     z+81z=30 or z2−30z+81=0 or     (z−27)(z−3)=0 i.e.,     81sin2⁡θ=34sin2⁡θ=33 or 31∴     sin2⁡θ=34,14 or sin⁡θ=±32,±12∴     θ=30∘,60∘,120∘,150∘
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