Solving trig equations and the unit circle

Worked solutions

Problem 1

1. $\sin\tfrac\pi6 = \tfrac12$, so the reference angle is $\tfrac\pi6$.From the table: $\tfrac\pi6$ is the quadrant-I angle whose sine is $\tfrac12$.
2. $x = \tfrac\pi6$ or $x = \pi - \tfrac\pi6 = \tfrac{5\pi}6$.$\tfrac12 > 0$, and sine is positive in quadrants I and II only, so there is one solution in each. The quadrant-II angle $\pi - \tfrac\pi6$ is the mirror image of $\tfrac\pi6$ across the vertical axis, at the same height, so it has the same sine.
3. Every solution is $\tfrac\pi6$ or $\tfrac{5\pi}6$ plus a whole number of turns.Sine has period $2\pi$: adding a full turn returns to the same point of the circle, and $[0, 2\pi)$ is exactly one turn, so every angle is one of its points plus $2k\pi$.
4. $x = \tfrac\pi6, \tfrac{5\pi}6$; general $x = \tfrac\pi6 + 2k\pi$ or $x = \tfrac{5\pi}6 + 2k\pi$Steps 2 and 3.

Problem 2

1. $\cos\tfrac\pi6 = \tfrac{\sqrt3}2$, so the reference angle is $\tfrac\pi6$.The table gives the value without its sign; the quadrant rule supplies the sign.
2. The solutions lie in quadrants II and III.$-\tfrac{\sqrt3}2 < 0$, and cosine is positive in quadrants I and IV, so it is negative in II and III.
3. $x = \pi - \tfrac\pi6 = \tfrac{5\pi}6$ and $x = \pi + \tfrac\pi6 = \tfrac{7\pi}6$.These are the quadrant-II and quadrant-III angles with reference angle $\tfrac\pi6$. Both lie at horizontal coordinate $-\cos\tfrac\pi6 = -\tfrac{\sqrt3}2$.
4. $x = \tfrac{5\pi}6, \tfrac{7\pi}6$Step 3.

Problem 3

1. Let $s = \sin x$; the equation is $2s^2 - s - 1 = 0$.Only $\sin x$ appears, so this is a quadratic in the one number $s$.
2. $2s^2 - s - 1 = (2s + 1)(s - 1)$.Expanding the right side gives $2s^2 - 2s + s - 1$, which is the left side.
3. $\sin x = 1$ or $\sin x = -\tfrac12$.A product is zero exactly when one of its factors is.
4. $\sin x = 1$ gives $x = \tfrac\pi2$ only.The unit circle reaches height 1 at one point, the top, $(0, 1)$. Its quadrant-II partner $\pi - \tfrac\pi2$ is $\tfrac\pi2$ again, so this value has one solution per turn, not two.
5. $\sin x = -\tfrac12$ gives $x = \pi + \tfrac\pi6 = \tfrac{7\pi}6$ and $x = 2\pi - \tfrac\pi6 = \tfrac{11\pi}6$.The reference angle is $\tfrac\pi6$ (Problem 1), and $-\tfrac12 < 0$, so the solutions lie in quadrants III and IV, where sine is negative.
6. $x = \tfrac\pi2, \tfrac{7\pi}6, \tfrac{11\pi}6$Steps 4 and 5 together, in increasing order.

Problem 4

1. $2\sin x\cos x = \cos x$.The double-angle formula writes both sides in terms of the same angle $x$.
2. $2\sin x\cos x - \cos x = \cos x\,(2\sin x - 1) = 0$.Move everything to one side and factor out $\cos x$. Dividing both sides by $\cos x$ instead would assume $\cos x \neq 0$ and throw away every solution where $\cos x = 0$ (mistake 1); as a factor, $\cos x = 0$ stays one of the cases.
3. $\cos x = 0$ gives $x = \tfrac\pi2, \tfrac{3\pi}2$.Cosine is the horizontal coordinate, which is 0 at the top and the bottom of the circle.
4. $2\sin x - 1 = 0$, that is $\sin x = \tfrac12$, gives $x = \tfrac\pi6, \tfrac{5\pi}6$.Problem 1.
5. $x = \tfrac\pi6, \tfrac\pi2, \tfrac{5\pi}6, \tfrac{3\pi}2$The two cases together, in increasing order. At $\tfrac\pi2$: $\sin\pi = 0 = \cos\tfrac\pi2$.

Problem 5

1. $\tan\tfrac\pi3 = \dfrac{\sin(\pi/3)}{\cos(\pi/3)} = \dfrac{\sqrt3/2}{1/2} = \sqrt3$, so the reference angle is $\tfrac\pi3$.$\tan x = \sin x/\cos x$, with both values from the table.
2. $x = \tfrac\pi3$ or $x = \pi + \tfrac\pi3 = \tfrac{4\pi}3$.$\sqrt3 > 0$, and tangent is positive in quadrants I and III, where sine and cosine have the same sign.
3. $\tan(x + \pi) = \tan x$ wherever $\tan x$ is defined.Adding $\pi$ moves $(\cos x, \sin x)$ to the opposite point $(-\cos x, -\sin x)$, and negating both leaves their ratio unchanged. So tangent has period $\pi$, and the two solutions in step 2 are one solution and its shift by $\pi$.
4. $x = \tfrac\pi3, \tfrac{4\pi}3$; general $x = \tfrac\pi3 + k\pi$Every solution is $\tfrac\pi3$ plus a whole number of periods; even $k$ gives $\tfrac\pi3$ plus whole turns, odd $k$ gives $\tfrac{4\pi}3$ plus whole turns.

Problem 6

1. $2\cos^2x - 1 = \cos x$.Of the three forms of $\cos 2x$, $2\cos^2x - 1$ is the one with only cosines, so the equation becomes one in $\cos x$ alone.
2. $2\cos^2x - \cos x - 1 = (2\cos x + 1)(\cos x - 1) = 0$.Move everything to one side and factor the quadratic in $\cos x$; expanding gives $2\cos^2x - 2\cos x + \cos x - 1$. Factoring finds every solution, because a product is zero exactly when one factor is; nothing is divided, so no case is lost.
3. $\cos x = 1$ gives $x = 0$.The circle's horizontal coordinate is 1 at one point only, $(1, 0)$, which is the angle 0; $2\pi$ is the same point but is outside $[0, 2\pi)$.
4. $\cos x = -\tfrac12$ gives $x = \pi - \tfrac\pi3 = \tfrac{2\pi}3$ and $x = \pi + \tfrac\pi3 = \tfrac{4\pi}3$.$\cos\tfrac\pi3 = \tfrac12$ gives the reference angle, and $-\tfrac12 < 0$ puts the solutions in quadrants II and III, where cosine is negative.
5. $x = 0, \tfrac{2\pi}3, \tfrac{4\pi}3$Steps 3 and 4, in increasing order. At $\tfrac{2\pi}3$: $\cos\tfrac{4\pi}3 = -\tfrac12 = \cos\tfrac{2\pi}3$.

Problem 7

1. At any solution, $\cos x \neq 0$.If $\cos x = 0$, the equation would force $\sin x = 0$ as well, but $\sin^2x + \cos^2x = 1$ rules out both being 0. So dividing by $\cos x$ loses nothing here, unlike Problem 4, where $\cos x = 0$ did give solutions.
2. $\tan x = 1$.Divide both sides by $\cos x$, nonzero at every solution by step 1.
3. $x = \tfrac\pi4$ or $x = \pi + \tfrac\pi4 = \tfrac{5\pi}4$.$\sin\tfrac\pi4 = \cos\tfrac\pi4 = \tfrac{\sqrt2}2$, so $\tan\tfrac\pi4 = 1$ and $\tfrac\pi4$ is the reference angle; $1 > 0$, and tangent is positive in quadrants I and III.
4. $x = \tfrac\pi4, \tfrac{5\pi}4$Step 3.

Problem 8

1. As $x$ runs over $[0, 2\pi)$, $3x$ runs over $[0, 6\pi)$: three full turns.Multiplying by 3 stretches the interval to three times its length.
2. There are six solutions.$\cos 3x = \tfrac12$, and cosine takes the value $\tfrac12$ twice in each turn (once in quadrant I and once in IV, since $\tfrac12$ is strictly between $-1$ and 1), so three turns give $3 \times 2 = 6$.
3. $\cos 3x = \tfrac12$, so $3x = \pm\tfrac\pi3 + 2k\pi$.Divide by 2. $\cos\tfrac\pi3 = \tfrac12$ gives the reference angle; cosine is positive in quadrants I and IV, and the quadrant-IV solution $2\pi - \tfrac\pi3$ is $-\tfrac\pi3$ plus a turn. Cosine has period $2\pi$.
4. $3x = \tfrac\pi3, \tfrac{5\pi}3, \tfrac{7\pi}3, \tfrac{11\pi}3, \tfrac{13\pi}3, \tfrac{17\pi}3$.The values of step 3 in $[0, 6\pi)$: $\tfrac\pi3 + 2k\pi$ for $k = 0, 1, 2$ and $-\tfrac\pi3 + 2k\pi$ for $k = 1, 2, 3$.
5. $x = \tfrac\pi9, \tfrac{5\pi}9, \tfrac{7\pi}9, \tfrac{11\pi}9, \tfrac{13\pi}9, \tfrac{17\pi}9$Divide by 3. Six solutions, as step 2 predicted.

Problem 9

1. $\sin x + \cos x = \sqrt2\big(\sin x\cos\tfrac\pi4 + \cos x\sin\tfrac\pi4\big) = \sqrt2\sin(x + \tfrac\pi4)$.$\sqrt2\cos\tfrac\pi4 = \sqrt2\sin\tfrac\pi4 = 1$, and the bracket is the sum formula with the upper sign. One sine is easier to solve than a sum of two.
2. $\sin(x + \tfrac\pi4) = \tfrac1{\sqrt2} = \tfrac{\sqrt2}2$.Divide the equation $\sqrt2\sin(x + \tfrac\pi4) = 1$ by $\sqrt2$.
3. $x + \tfrac\pi4 = \tfrac\pi4$ or $\tfrac{3\pi}4$.The reference angle is $\tfrac\pi4$ and sine is positive in quadrants I and II, giving $\tfrac\pi4$ and $\pi - \tfrac\pi4$. As $x$ runs over $[0, 2\pi)$, $x + \tfrac\pi4$ runs over $[\tfrac\pi4, \tfrac{9\pi}4)$; the next solution, $\tfrac\pi4 + 2\pi$, is just outside it.
4. $x = 0$ or $x = \tfrac\pi2$.Subtract $\tfrac\pi4$. At 0: $0 + 1 = 1$; at $\tfrac\pi2$: $1 + 0 = 1$.
5. Squaring instead: $\sin^2x + 2\sin x\cos x + \cos^2x = 1$, so $1 + \sin 2x = 1$ and $\sin 2x = 0$.Expand the square; $\sin^2x + \cos^2x = 1$ and $2\sin x\cos x = \sin 2x$.
6. $2x = 0, \pi, 2\pi, 3\pi$, so the candidates are $x = 0, \tfrac\pi2, \pi, \tfrac{3\pi}2$.Sine is 0 at the multiples of $\pi$, and $2x$ runs over $[0, 4\pi)$.
7. At $\pi$: $\sin\pi + \cos\pi = 0 - 1 = -1$. At $\tfrac{3\pi}2$: $-1 + 0 = -1$.Substitute each candidate into the original equation. Squaring cannot tell $1$ from $-1$, since both square to 1, so the squared equation also holds wherever $\sin x + \cos x = -1$; these two candidates are those points.
8. $x = 0, \tfrac\pi2$; squaring gives $1 + \sin 2x = 1$, i.e. $\sin 2x = 0$, whose solutions $0, \tfrac\pi2, \pi, \tfrac{3\pi}2$ include two extraneous ones ($\pi$ and $\tfrac{3\pi}2$ give $-1$)Steps 4 and 7: of the four candidates, only 0 and $\tfrac\pi2$ satisfy the original equation, as step 4 found without squaring.

Problem 10

1. $\sin x = \tfrac{\sqrt3}2$ or $\sin x = -\tfrac{\sqrt3}2$.The numbers whose square is $\tfrac34$ are $\pm\tfrac{\sqrt3}2$. Keeping both signs loses nothing and adds nothing, so unlike squaring in Problem 9 this step is reversible.
2. On $[0, 2\pi)$: $x = \tfrac\pi3, \tfrac{2\pi}3, \tfrac{4\pi}3, \tfrac{5\pi}3$.$\sin\tfrac\pi3 = \tfrac{\sqrt3}2$ gives the reference angle. The plus sign gives quadrants I and II ($\tfrac\pi3$, $\pi - \tfrac\pi3$), the minus sign quadrants III and IV ($\pi + \tfrac\pi3$, $2\pi - \tfrac\pi3$): one solution in every quadrant.
3. $\sin^2(x + \pi) = \sin^2x$.$\sin(x + \pi) = -\sin x$, because adding $\pi$ moves to the opposite point of the circle, and squaring removes the sign. So $\sin^2x$ has period $\pi$, not $2\pi$.
4. $\tfrac{4\pi}3 = \tfrac\pi3 + \pi$, $\tfrac{2\pi}3 = -\tfrac\pi3 + \pi$ and $\tfrac{5\pi}3 = -\tfrac\pi3 + 2\pi$.By step 3, the four solutions of step 2 are $\tfrac\pi3$ and $-\tfrac\pi3$ shifted by whole periods $\pi$.
5. $x = \pm\tfrac\pi3 + k\pi$ (on $[0, 2\pi)$: $\tfrac\pi3, \tfrac{2\pi}3, \tfrac{4\pi}3, \tfrac{5\pi}3$)Steps 3 and 4: every solution is $\tfrac\pi3$ or $-\tfrac\pi3$ plus a whole number of periods.