If 0°<θ<90°
2sin²θ + 3cosθ = 3, then the value of θ is ???

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2 thoughts on “If 0°<θ<90°<br />2sin²θ + 3cosθ = 3, then the value of θ is ???<br /><br />GUD wali morning ☺❤❣️✌​”

1. Answe accha

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2. Answer:

   Required measure of angle A is 90°.

   Step-by-step explanation:

   = > cos^2 A – 3 cosA + 2 = 2 sin^2 A

   From the properties of trigonometry :

   sin^2 theta + cos^2 theta= 1

   sin^2 theta= 1 – cos^2 theta

   theta will be cancelled

   = > cos^2 A – 3 cosA + 2 = 2( 1 – cos^2 A ) { 2sin^2 A = 2( 1 – cos^2 A ) }

   = > cos^2 A – 3 cosA + 2 = 2 – 2 cos^2 A

   = > 2 cos^2 A + cos^2 A – 3 cosA + 2 – 2 = 0

   = > 3 cos^2 A – 3 cosA = 0

   = > 3 cosA ( cosA – 1 ) = 0

   Case 1 : If cosA is 0 .

   = > 3 cosA = 0

   = > cosA = 0

   = > cosA = cos90°

   = > A = 90° or π / 2

   Case 2 : If cosA – 1 is zero.

   = > cosA – 1 = 0

   = > cosA = 1

   = > cosA = cos0°

   But here, theta or A is greater than 0°, cosA ≠ cos0°

   Therefore the required measure of angle A is 90°.

   This is your answer

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