Question

When 0°<A<90°, solve the equation: 2cos²A+sin A-2=0

Answer 1

Given,

⇒2 cos²A + sin A - 2 = 0

Using identity,

⇒cos²A = ( 1 - sin²A )

⇒ 2( 1 - sin²A ) + sin A - 2 = 0

⇒ 2 - 2 sin²A + sin A - 2 = 0

⇒ - 2 sin²A + sin A = 0

⇒ sin A = 2 sin²A

⇒( sin A ) / 2 = sin²A

We shall solve it by trial and error method,

Let , A = 30°

⇒ ( sin A ) / 2 = sin²A

⇒ ( sin 30° ) / 2 = ( sin 30° )²

⇒( 1/2 ) / 2 = ( 1/2 )² [ sin 30° = ( 1/2 ) ]

⇒ ( 1/4 ) = ( 1/4 )

Hence, A = 30°.

Answer 2

Answer:

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Step-by-step explanation:

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