If 2sin²Beta-cos²Beta =2 then Beta is
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SOLUTION
TO DETERMINE
If 2sin²β – cos²β = 2 then β is
EVALUATION
Here it is given that
2sin²β – cos²β = 2
Now
2sin²β – cos²β = 2
⇒ 2sin²β – ( 1 - sin²β ) = 2
⇒ 2sin²β - 1 + sin²β = 2
⇒ 3sin²β - 1 = 2
⇒ 3sin²β = 3
⇒ sin²β = 1
⇒ sinβ = 1
⇒ β = 90°
FINAL ANSWER
If 2sin²β – cos²β = 2 then β is 90°
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