\( \sin \theta+\cos \theta=0 \) \( \sin \left(3 x+30^{\circ}\right)=-\cos 2 x \) \( 2 \sin ^{2} x+\sin x=3 \) \( 1+\sin \theta=\cos ^{2} \theta \) \( 2 \cos ^{2} x=\cos x \) Consider the identity \( \frac{1+2 \sin x \cdot \cos x}{\sin x+\cos x}=1 \) 1 For which values of \( x \) will the identity be undefing 2 State the general solution of \( \frac{1+2 \sin x \cdot \cos x}{\sin x+\cos x}=1 \)

We begin with the identity   (1 + 2 sin x · cos x)⁄(sin x + cos x) = 1. Notice that the numerator can be rewritten using a well‐known trigonometric identity. Since   sin²x + cos²x = 1,   and (sin x + cos x)² = sin²x + 2 sin x cos x + cos²x,   we have   (sin x + cos x)² = 1 + 2 sin x cos x. Thus, the left‐hand side becomes   (sin x + cos x)²⁄(sin x + cos x). However, this cancellation (to leave sin x + cos x on the top) is valid only when the denominator is not zero. Our work will now proceed in two parts. ────────────────────────────── 1. For which values of x is the expression undefined? The expression is undefined when its denominator is zero, that is when   sin x + cos x = 0. To solve this, write   sin x = - cos x            (provided cos x ≠ 0).   Dividing by cos x gives   tan x = -1. The general solution for tan x = -1 is   x = -π/4 + πk             where k is any integer. (Equivalently, one may also write x = 3π/4 + πk, since these two forms represent the same set of angles.) ────────────────────────────── 2. State the general solution of   (1 + 2 sin x cos x)/(sin x + cos x) = 1. For all x such that sin x + cos x ≠ 0, we may cancel one factor of (sin x + cos x) to obtain   sin x + cos x = 1. This simplified equation must now be solved. A useful step is to express sin x + cos x as a single sine function. Recall that   sin x + cos x = √2 · sin(x + π/4). Thus the equation becomes   √2 · sin(x + π/4) = 1,   or sin(x + π/4) = 1/√2. We know that sin u = 1/√2 when   u = π/4 + 2πk          or  u = 3π/4 + 2πk   (k an integer). Substitute back u = x + π/4: • For u = π/4 + 2πk:   x + π/4 = π/4 + 2πk ⟹ x = 2πk. • For u = 3π/4 + 2πk:   x + π/4 = 3π/4 + 2πk ⟹ x = π/2 + 2πk. Thus, the general solution of sin x + cos x = 1 is   x = 2πk   or  x = π/2 + 2πk     (for any integer k). Before finalizing our answer, we should recall that these solutions are valid only when the original denominator is nonzero. Checking the denominator for these values: • For x = 2πk:   sin(2πk) + cos(2πk) = 0 + 1 = 1 ≠ 0. • For x = π/2 + 2πk:   sin(π/2 + 2πk) + cos(π/2 + 2πk) = 1 + 0 = 1 ≠ 0. So the solutions are indeed acceptable. ────────────────────────────── Final Answers: 1. The expression is undefined when sin x + cos x = 0, i.e., when   x = -π/4 + πk   (or equivalently x = 3π/4 + πk), with k any integer. 2. For the identity   (1 + 2 sin x cos x)/(sin x + cos x) = 1 to hold (with sin x + cos x ≠ 0), the general solution is   x = 2πk   or  x = π/2 + 2πk,  with k any integer.