When solving 2sin²x = sinx, a student divides both sides by sinx and gets 2sinx = 1, arriving at x = π/6 and x = 5π/6 in [0, 2π). What is wrong with this approach?
ANothing is wrong — dividing by sinx is the standard algebraic technique
BThe student should have used the Pythagorean identity instead of dividing
CThe division by sinx discards solutions where sinx = 0, namely x = 0 and x = π
DThe equation has no solutions in [0, 2π) because both sides are quadratic
Dividing both sides by sinx assumes sinx ≠ 0 and discards any solutions where sinx = 0. Moving everything to one side and factoring is safe: 2sin²x − sinx = 0 → sinx(2sinx − 1) = 0, giving sinx = 0 (yielding x = 0, π) or 2sinx − 1 = 0 → sinx = 1/2 (yielding x = π/6, 5π/6). There are four solutions total, not two. Whenever you divide by an expression that might equal zero, you risk losing solutions.
Question 2 Multiple Choice
What is the general solution to sin(x) = 1/2?
Ax = π/6 only, since arcsin(1/2) = π/6
Bx = π/6 + πn for all integers n
Cx = π/6 + 2πn and x = 5π/6 + 2πn for all integers n
Dx = π/6 + 2πn for all integers n
The inverse sine function gives only one value (arcsin(1/2) = π/6), but the unit circle shows two angles in [0, 2π) where sin equals 1/2: π/6 (quadrant I) and π/6's symmetric partner 5π/6 (quadrant II), since sin is positive in both. Then periodicity adds 2πn to each branch. Option B (adding πn) confusingly mixes the period of sin (2π) with the spacing between the two solutions; option D forgets the second branch at 5π/6.
Question 3 True / False
When solving a trigonometric equation, applying the Pythagorean identity (sin²x + cos²x = 1) can convert an equation involving both sine and cosine into an equation in a single trig function, which can then be solved by factoring.
TTrue
FFalse
Answer: True
This is a key technique for equations that mix trig functions. For example, 1 − cos²x = sinx becomes sin²x = sinx (via the Pythagorean identity), which factors as sinx(sinx − 1) = 0. The identity converts the two-function problem into a single-function problem suitable for standard solving techniques.
Question 4 True / False
The equation cos(x) = 0.8 has exactly two solutions: x = arccos(0.8) and x = −arccos(0.8).
TTrue
FFalse
Answer: False
These are the two solutions in (−π, π], but because cosine is periodic with period 2π, there are infinitely many solutions. The general solution is x = ±arccos(0.8) + 2πn for all integers n. Inverse trig functions return only one value (within their restricted range) precisely so that they are functions — but this means you must always add the periodicity term to capture all solutions.
Question 5 Short Answer
Why is factoring the correct approach when solving sin(x)·cos(x) = sin(x), rather than dividing both sides by sin(x)? What solutions would be lost?
Think about your answer, then reveal below.
Model answer: Dividing by sin(x) assumes sin(x) ≠ 0, which discards any solutions where sin(x) = 0 (i.e., x = 0, π, 2π, … in [0, 2π]). The correct approach is to move everything to one side: sin(x)·cos(x) − sin(x) = 0, then factor: sin(x)(cos(x) − 1) = 0. Setting each factor to zero gives sin(x) = 0 or cos(x) = 1. Both cases must be solved, yielding x = 0, π (from sin = 0) and x = 0 (from cos = 1). Factoring finds all solutions; division hides the ones that make the divisor zero.
This is the most dangerous shortcut in solving trig equations. Division by a trig expression is only valid when you can prove that expression is never zero on the domain — which is rarely true. Factoring is always safe because zero product property doesn't throw away cases; it just separates them into individual equations.