The angle A lies in the third quadrant and it satisfies the equation 4

1.	The angle A lies in the third quadrant and it satisfies the equation 4 (sin2x + cos x) = 1. What is the measure of angle A?
Answer» 4 sin²x + 4 cos \(x\) = 1 ⇒ 4 sin²x + 4 cos \(x\) – 1 = 0 ⇒ 4 (1 – cos²x) + 4 cos \(x\) – 1 = 0 ⇒ – 4 cos²x + 4 cos x + 3 = 0 ⇒ 4 cos²x – 4 cos \(x\) – 3 = 0 ⇒ 4 cos²x – 6 cos \(x\) + 2 cos \(x\) – 3 = 0 ⇒ 2 cos \(x\) (2 cos \(x\) – 3) + 1 (2 cos \(x\) – 3) = 0 ⇒ (2 cos \(x\) + 1) (2 cos \(x\) – 3) = 0 ⇒ cos \(x\) = \(-\frac12\) and cos \(x\) = \(\frac32\) (not possible) Now cos \(x\) = \(-\frac12\) Since A lies in the third quadrant and cos A = \(-\frac12\), therefore, cos A = cos (180º + 60º) = cos 240º ⇒ A = 240º.

The angle A lies in the third quadrant and it satisfies the equation 4 (sin2x + cos x) = 1. What is the measure of angle A?

Answer»

4 sin²x + 4 cos \(x\) = 1

⇒ 4 sin²x + 4 cos \(x\) – 1 = 0 ⇒ 4 (1 – cos²x) + 4 cos \(x\) – 1 = 0

⇒ – 4 cos²x + 4 cos x + 3 = 0 ⇒ 4 cos²x – 4 cos \(x\) – 3 = 0

⇒ 4 cos²x – 6 cos \(x\) + 2 cos \(x\) – 3 = 0 ⇒ 2 cos \(x\) (2 cos \(x\) – 3) + 1 (2 cos \(x\) – 3) = 0

⇒ (2 cos \(x\) + 1) (2 cos \(x\) – 3) = 0

⇒ cos \(x\) = \(-\frac12\) and cos \(x\) = \(\frac32\) (not possible)

Now cos \(x\) = \(-\frac12\)

Since A lies in the third quadrant and cos A = \(-\frac12\), therefore,

cos A = cos (180º + 60º) = cos 240º ⇒ A = 240º.