Q.29 Which one of the following is the solution for
cos²x + 2cosx + 1 = 0, for values of x in the range of
0° < x < 360°?

(A) 45°
(B) 90°
(C) 180°
(D) 270°

Introduction to Solving cos²x + 2cosx + 1 = 0

The trigonometric equation cos²x + 2cosx + 1 = 0 appears in exams testing quadratic trig identities. Substitute y = cosx to get y² + 2y + 1 = 0, or (y + 1)² = 0, so y = -1 (double root). Thus, cosx = -1 only at x = 180° in 0° < x < 360°.

Solving the Equation

Recognize that:

cos²x + 2cosx + 1 = (cosx + 1)² = 0

So:

cosx + 1 = 0

cosx = -1

In the range 0° < x < 360°, cosine equals -1 only at:

x = 180°

Step-by-Step Solution Process

Let y = cosx. The equation becomes:

y² + 2y + 1 = 0

Factor:

(y + 1)² = 0

Therefore:

y = -1

General solution: x = (2k + 1)180°, for integer k.

In 0° < x < 360°, only x = 180° fits.

Option Analysis

45°: cos45° = √2/2 ≈ 0.707 ⇒ (cosx + 1)² > 0, not zero.

90°: cos90° = 0 ⇒ (0 + 1)² = 1 ≠ 0.

180°: cos180° = -1 ⇒ (-1 + 1)² = 0 ✔ satisfies.

270°: cos270° = 0 ⇒ (0 + 1)² = 1 ≠ 0.

Verifying Each Multiple-Choice Option

Final Answer

x = 180° is the ONLY solution in the interval 0° < x < 360°.