Solving Trigonometric Equations

The graph of y=\cos x for 0^\circ to 360^\circ is symmetrical about x=180^\circ.

1. The cos graph is symmetrical about x=180^\circ
2. So the other solution is 360^\circ-134.4^\circ
3. =225.6^\circ
4. (45.6^\circ uses the sine symmetry 180^\circ-x, 314.4^\circ adds 180^\circ, which is the tan rule, and -134.4^\circ is outside the range.)

Your calculator gives one solution; the cos graph is symmetrical about x=180^\circ.

1. \cos^{-1}0.28=73.7^\circ
2. The cos graph is symmetrical about 180^\circ, so the other solution is 360^\circ-73.7^\circ
3. =286.3^\circ
4. Answer: x=73.7^\circ and x=286.3^\circ

Divide by 7 first; the sin graph is symmetrical about x=90^\circ.

1. \sin x=\dfrac27
2. \sin^{-1}\left(\dfrac27\right)=16.6^\circ
3. The other solution is 180^\circ-16.6^\circ=163.4^\circ
4. Answer: x=16.6^\circ and x=163.4^\circ

Your calculator gives a negative angle; the tan graph repeats every 180^\circ, so keep adding 180^\circ.

1. \tan x=-\dfrac74=-1.75
2. \tan^{-1}(-1.75)=-60.3^\circ, which is outside the range
3. tan repeats every 180^\circ
4. -60.3^\circ+180^\circ=119.7^\circ
5. 119.7^\circ+180^\circ=299.7^\circ
6. x=119.7^\circ and x=299.7^\circ

Your calculator gives a negative angle, which is outside the range: sin is negative between 180^\circ and 360^\circ.

1. \sin x=-\dfrac23
2. \sin^{-1}\left(-\dfrac23\right)=-41.8^\circ, which is outside the range
3. sin is negative between 180^\circ and 360^\circ
4. 180^\circ+41.8^\circ=221.8^\circ
5. 360^\circ-41.8^\circ=318.2^\circ
6. Answer: x=221.8^\circ and x=318.2^\circ

Rearrange to get \cos x on its own, then use the exact value \cos45^\circ=\dfrac{1}{\sqrt2}.

1. \sqrt2\cos x=-1, so \cos x=-\dfrac{1}{\sqrt2}
2. \cos45^\circ=\dfrac{1}{\sqrt2}, and cos is negative between 90^\circ and 270^\circ
3. 180^\circ-45^\circ=135^\circ
4. 180^\circ+45^\circ=225^\circ
5. Answer: x=135^\circ and x=225^\circ

Use the symmetry of the graph of y=\sin x: \sin x is negative between 180^\circ and 360^\circ.

1. \sin x=-k is negative, so x is between 180^\circ and 360^\circ
2. By the symmetry of the sin graph, \sin(180^\circ+28^\circ)=-\sin28^\circ
3. and \sin(360^\circ-28^\circ)=-\sin28^\circ
4. Answer: x=208^\circ and x=332^\circ

Rearrange so that \sin x and \cos x are on opposite sides, then divide by \cos x and use \tan x=\dfrac{\sin x}{\cos x}.

1. 5\sin x=-2\cos x
2. Divide both sides by 5\cos x: \dfrac{\sin x}{\cos x}=-\dfrac25
3. \tan x=-0.4
4. \tan^{-1}(-0.4)=-21.8^\circ, which is outside the range
5. tan repeats every 180^\circ: -21.8^\circ+180^\circ=158.2^\circ
6. 158.2^\circ+180^\circ=338.2^\circ
7. Answer: x=158.2^\circ and x=338.2^\circ

Find the two solutions between 0^\circ and 360^\circ first; the sin graph repeats every 360^\circ.

1. \sin x=\dfrac13
2. \sin^{-1}\left(\dfrac13\right)=19.5^\circ
3. 180^\circ-19.5^\circ=160.5^\circ
4. Add 360^\circ to each: 379.5^\circ and 520.5^\circ
5. Answer: x=19.5^\circ,\ 160.5^\circ,\ 379.5^\circ,\ 520.5^\circ

Divide both sides by \sqrt3 and simplify: \dfrac{3}{\sqrt3} is a simpler surd.

1. \tan x=-\dfrac{3}{\sqrt3}
2. \dfrac{3}{\sqrt3}=\dfrac{3\sqrt3}{3}=\sqrt3, so \tan x=-\sqrt3
3. \tan60^\circ=\sqrt3, so one solution is x=-60^\circ
4. tan repeats every 180^\circ: -60^\circ+180^\circ=120^\circ
5. -60^\circ-180^\circ=-240^\circ is outside the range
6. Answer: x=-60^\circ and x=120^\circ

When you take the square root, remember the negative root as well as the positive one.

1. \cos^2 x=\dfrac25=0.4
2. \cos x=\sqrt{0.4}=0.632\ldots or \cos x=-\sqrt{0.4}=-0.632\ldots
3. \cos x=0.632\ldots: x=50.8^\circ or 360^\circ-50.8^\circ=309.2^\circ
4. \cos x=-0.632\ldots: x=129.2^\circ or 360^\circ-129.2^\circ=230.8^\circ
5. Answer: x=50.8^\circ,\ 129.2^\circ,\ 230.8^\circ,\ 309.2^\circ

Square root both sides, keeping both the positive and the negative root, then only give the angles in the range.

1. \sin^2 x=\dfrac34
2. \sin x=\dfrac{\sqrt3}{2} or \sin x=-\dfrac{\sqrt3}{2}
3. \sin x=\dfrac{\sqrt3}{2} gives 60^\circ and 120^\circ, which are both outside the range
4. \sin x=-\dfrac{\sqrt3}{2}: sin is negative between 180^\circ and 360^\circ
5. 180^\circ+60^\circ=240^\circ and 360^\circ-60^\circ=300^\circ
6. Answer: x=240^\circ and x=300^\circ

Factorise: dividing by \tan x would lose some solutions.

1. \tan x(2\tan x+5)=0
2. \tan x=0 or \tan x=-2.5
3. \tan x=0: x=180^\circ (0^\circ and 360^\circ are not in the range 0^\circ<x<360^\circ)
4. \tan x=-2.5: \tan^{-1}(-2.5)=-68.2^\circ
5. so x=-68.2^\circ+180^\circ=111.8^\circ or 111.8^\circ+180^\circ=291.8^\circ
6. Answer: x=111.8^\circ,\ 180^\circ,\ 291.8^\circ

Treat it as a quadratic in \sin x and factorise; check whether each value of \sin x is possible.

1. Let s=\sin x: 4s^2+7s-2=0
2. (4s-1)(s+2)=0
3. \sin x=\dfrac14 or \sin x=-2
4. \sin x=-2 has no solutions, because -1\leqslant\sin x\leqslant1
5. \sin x=\dfrac14: x=14.5^\circ or 180^\circ-14.5^\circ=165.5^\circ
6. Answer: x=14.5^\circ and x=165.5^\circ

Treat it as a quadratic in \cos x: factorise, then find two angles for each value of \cos x.

1. Let c=\cos x: 6c^2+c-2=0
2. (2c-1)(3c+2)=0
3. \cos x=\dfrac12 or \cos x=-\dfrac23
4. \cos x=\dfrac12: x=60^\circ or 360^\circ-60^\circ=300^\circ
5. \cos x=-\dfrac23: x=131.8^\circ or 360^\circ-131.8^\circ=228.2^\circ
6. Answer: x=60^\circ,\ 131.8^\circ,\ 228.2^\circ,\ 300^\circ

Factorise the numerator and use \sin^2 x+\cos^2 x=1, then \tan x=\dfrac{\sin x}{\cos x}. Answer part (a) before attempting part (b).

Part (a)

1. 2-2\cos^2 x=2(1-\cos^2 x)
2. 1-\cos^2 x=\sin^2 x, so the numerator is 2\sin^2 x
3. f(x)=\dfrac{2\sin^2 x}{\sin x\cos x}=\dfrac{2\sin x}{\cos x}
4. Answer: 2\tan x

Part (b)

1. 2\tan x=-2, so \tan x=-1
2. \tan45^\circ=1, so \tan x=-1 when x=180^\circ-45^\circ=135^\circ
3. tan repeats every 180^\circ: 135^\circ+180^\circ=315^\circ
4. Answer: x=135^\circ and x=315^\circ

Use area =\dfrac12ab\sin C to find \sin P; an angle in a triangle can be acute or obtuse. Answer part (a) before attempting part (b).

Part (a)

1. Area =\dfrac12\times PQ\times PR\times\sin P
2. \dfrac12\times9\times6\times\sin P=20
3. 27\sin P=20, so \sin P=\dfrac{20}{27}
4. \sin^{-1}\left(\dfrac{20}{27}\right)=47.8^\circ
5. The angle in a triangle can also be obtuse: 180^\circ-47.8^\circ=132.2^\circ
6. Answer: 47.8^\circ and 132.2^\circ

Part (b)

1. Use the cosine rule with angle P=132.2^\circ
2. QR^2=9^2+6^2-2\times9\times6\times\cos132.2^\circ
3. QR^2=117+72.55\ldots=189.55\ldots
4. QR=\sqrt{189.55\ldots}
5. Answer: QR=13.8 cm

Use \tan\theta=\dfrac{\sin\theta}{\cos\theta} to find \tan\theta, and \sin^2\theta+\cos^2\theta=1 to find k. Answer part (a) before attempting part (b).

Part (a)

1. \tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{2k}{-3k}
2. Answer: \tan\theta=-\dfrac23

Part (b)

1. \tan^{-1}\left(-\dfrac23\right)=-33.7^\circ, which is not between 90^\circ and 180^\circ
2. tan repeats every 180^\circ: -33.7^\circ+180^\circ=146.3^\circ
3. Answer: \theta=146.3^\circ

Part (c)

1. \sin^2\theta+\cos^2\theta=1
2. (2k)^2+(-3k)^2=1, so 13k^2=1
3. k^2=\dfrac{1}{13} and k is positive
4. Answer: k=\dfrac{1}{\sqrt{13}}=\dfrac{\sqrt{13}}{13}

Use \sin^2 x+\cos^2 x=1 to replace \cos^2 x, so the equation only contains \sin x; then treat it as a quadratic.

Part (a)

1. Replace \cos^2 x with 1-\sin^2 x
2. 10(1-\sin^2 x)-\sin x=8
3. 10-10\sin^2 x-\sin x-8=0
4. 10\sin^2 x+\sin x-2=0
5. Factorise: (2\sin x+1)(5\sin x-2)=0
6. \sin x=-\dfrac12 or \sin x=\dfrac25

Part (b)

1. \sin x=\dfrac25: \sin^{-1}0.4=23.6^\circ
2. and 180^\circ-23.6^\circ=156.4^\circ
3. \sin x=-\dfrac12: \sin^{-1}(-0.5)=-30^\circ, which is outside the range
4. sin is negative between 180^\circ and 360^\circ: 180^\circ+30^\circ=210^\circ and 360^\circ-30^\circ=330^\circ
5. x=23.6^\circ,\ 156.4^\circ,\ 210^\circ,\ 330^\circ

Set the two expressions for y equal; the quadratic in \cos x does not factorise, so use the quadratic formula. Answer part (a) before attempting part (b).

Part (a)

1. At the points where the curves meet, 4\sin^2 x=2-3\cos x
2. Replace \sin^2 x with 1-\cos^2 x: 4-4\cos^2 x=2-3\cos x
3. Answer: 4\cos^2 x-3\cos x-2=0

Part (b)

1. Use the quadratic formula with \cos x as the unknown:
2. \cos x=\dfrac{3\pm\sqrt{(-3)^2-4\times4\times(-2)}}{2\times4}=\dfrac{3\pm\sqrt{41}}{8}
3. \cos x=1.175 or \cos x=-0.4254
4. \cos x=1.175 is impossible, because \cos x\leqslant1
5. \cos^{-1}(-0.4254)=115.2^\circ and 360^\circ-115.2^\circ=244.8^\circ
6. Answer: x=115.2^\circ and x=244.8^\circ