C2 Solving Trigonometric Equations | Maths Teaching - Free Printable

Let’s solve each equation one by one, carefully finding all solutions between 0° and 180°. We’ll use inverse trig functions and remember that sine, cosine, and tangent have different behaviors in different quadrants.

---

a) sin 2x = 0.5

We know sin θ = 0.5 when θ = 30° or 150° (in 0° to 360°).

So let 2x = 30° → x = 15°
Or 2x = 150° → x = 75°

But since x must be ≤ 180°, 2x can go up to 360°, so we check if there are more:

Next solution for sin θ = 0.5 is 30° + 360° = 390° → too big (since 2x ≤ 360°)

Wait — actually, within 0° to 360°, sin θ = 0.5 at 30° and 150° only? No — wait, sine is positive in Q1 and Q2, so yes, only those two in [0°, 360°].

But 2x can be up to 360°, so we’re good.

So x = 15°, 75°

Check: 2×15=30°, sin30=0.5 ✔️
2×75=150°, sin150=0.5 ✔️

Are there any others? What about 2x = 30° + 360° = 390° → x=195° > 180° ✘
Similarly, 2x=150+360=510→x=255>180✘

So only 15.0°, 75.0°

---

b) sin 3x = 1

sin θ = 1 when θ = 90° + 360°k

So 3x = 90° → x = 30°
Next: 3x = 450° → x = 150°
Next: 3x = 810° → x = 270° > 180° ✘

Check: 3×30=90°, sin90=1 ✔️
3×150=450°, sin450=sin(450-360)=sin90=1 ✔️

So 30.0°, 150°

---

c) cos 2x = 0.5

cos θ = 0.5 when θ = 60° or 300° (in 0° to 360°)

So 2x = 60° → x = 30°
2x = 300° → x = 150°

Check range: x ≤ 180°, so both OK.

Any more? Next would be 2x = 60+360=420→x=210>180✘
2x=300+360=660→x=330>180✘

So 30.0°, 150°

---

d) cos(x/2) = 0.5

Let θ = x/2 → cos θ = 0.5 → θ = 60° or 300° (but θ = x/2, and x ≤ 180° → θ ≤ 90°)

So θ = 60° → x = 120°
θ = 300° → x = 600° > 180° ✘

Also, cosine is even, but in [0°, 90°], only 60° works.

Wait — cos θ = 0.5 also at θ = -60°, but negative angles not considered here since x ≥ 0 → θ ≥ 0.

So only 120°

Check: x/2 = 60°, cos60=0.5 ✔️

---

e) tan 2x = 1

tan θ = 1 when θ = 45° + 180°k

So 2x = 45° → x = 22.5°
2x = 225° → x = 112.5°
2x = 405° → x = 202.5° > 180° ✘

Check: tan(45)=1, tan(225)=tan(180+45)=tan45=1 ✔️

So 22.5°, 112.5°

---

f) cos 2θ = -0.6

First, find reference angle: cos⁻¹(0.6) ≈ 53.13°

Cosine is negative in Q2 and Q3.

So 2θ = 180° - 53.13° = 126.87°
or 2θ = 180° + 53.13° = 233.13°

Then θ = 63.435° ≈ 63.4°
θ = 116.565° ≈ 117°

Check range: θ ≤ 180°, both OK.

Next solutions: 2θ = 126.87 + 360 = 486.87 → θ=243.4>180✘
2θ=233.13+360=593.13→θ=296.6>180✘

So 63.4°, 117°

---

g) tan 2θ = -0.5

tan θ = -0.5 → reference angle = tan⁻¹(0.5) ≈ 26.565°

Tangent is negative in Q2 and Q4.

So 2θ = 180° - 26.565° = 153.435° → θ ≈ 76.7°
or 2θ = 360° - 26.565° = 333.435° → θ ≈ 166.7°

Check: tan(153.435°) = tan(180-26.565) = -tan26.565 ≈ -0.5 ✔️
tan(333.435°) = tan(-26.565) ≈ -0.5 ✔️

Next: 2θ = 153.435 + 180 = 333.435 → already did
Then 2θ = 333.435 + 180 = 513.435 → θ=256.7>180✘

So 76.7°, 167° (rounded to 3 sig fig)

Wait: 166.7° rounds to 167°? Yes, 166.7 → 167 (3 sig fig)

Actually, 166.7 has 4 digits, but we want 3 significant figures: 167° is correct.

But let's compute precisely:

tan⁻¹(0.5) = arctan(0.5) ≈ 26.565051177°

So 2θ = 180 - 26.565051177 = 153.4349488° → θ = 76.7174744° ≈ 76.7°

2θ = 360 - 26.565051177 = 333.4349488° → θ = 166.7174744° ≈ 167°

Yes.

---

h) tan(θ/2) = √3

√3 ≈ 1.732, tan 60° = √3

So θ/2 = 60° → θ = 120°

Is there another? tan has period 180°, so next: θ/2 = 60° + 180° = 240° → θ = 480° > 180° ✘

Also, θ/2 must be ≤ 90° (since θ ≤ 180°), so only 60° works.

So 120°

Check: θ/2=60°, tan60=√3 ✔️

---

i) sin 4x = √2 / 2

√2 / 2 ≈ 0.7071, which is sin 45° and sin 135°

So 4x = 45° → x = 11.25°
4x = 135° → x = 33.75°
4x = 45° + 360° = 405° → x = 101.25°
4x = 135° + 360° = 495° → x = 123.75°
4x = 45° + 720° = 765° → x = 191.25° > 180° ✘
4x = 135° + 720° = 855° → x = 213.75° > 180° ✘

So possible x: 11.25°, 33.75°, 101.25°, 123.75°

All ≤ 180°, so all valid.

To 3 sig fig:
11.25° → 11.3°? Wait, 11.25 has 4 sig fig, but we need 3: 11.3°? Actually, 11.25 rounded to 3 sig fig is 11.3°? Let's see:

Significant figures: 11.25 — all digits are significant. To 3 sig fig: look at fourth digit (5), so round up: 11.3°

But 11.25 is exactly halfway? Standard rule: round to nearest even? But usually in such contexts, we just round 5 up.

Actually, 11.25 to 3 sig fig: the third digit is 2, fourth is 5 → round up 2 to 3 → 11.3°

Similarly, 33.75 → 33.8°
101.25 → 101° (since 101.25, third sig fig is 1, fourth is 2 <5 → 101°)
Wait, 101.25: digits are 1,0,1,2,5 — so first three are 101, and next is 2<5, so 101°

But 101 has three sig fig already.

Similarly, 123.75 → 124° (since 123.75, third digit 3, fourth 7>5 → round up to 124)

Let me list:

x = 11.25° → 11.3° (3 sig fig)
x = 33.75° → 33.8°
x = 101.25° → 101° (because 101.25, the '1' in hundreds, '0' tens, '1' units — that's three sig fig; decimal part doesn't add sig fig unless specified. Actually, 101.25 has five sig fig. To reduce to three: look at the first three digits: 101, and the next digit is 2<5, so 101°)

But 101° is exact? Or should we write 101.?

In context, probably acceptable as 101°.

Similarly, 123.75° → first three digits 124? 123.75: digits 1,2,3,7,5 — to three sig fig: look at fourth digit 7≥5, so round up 123 to 124 → 124°

But let's confirm values:

sin(4*11.25)=sin(45)=√2/2 ✔️
sin(4*33.75)=sin(135)=√2/2 ✔️
sin(4*101.25)=sin(405)=sin(45)=√2/2 ✔️
sin(4*123.75)=sin(495)=sin(495-360)=sin(135)=√2/2 ✔️

So answers: 11.3°, 33.8°, 101°, 124°

But 101.25 to 3 sig fig: since it's greater than 100, the third sig fig is the units place. 101.25 — the '1' is hundreds, '0' tens, '1' units — that's three sig fig. The .25 is extra. So to express with three sig fig, we can write 101° (implying uncertainty in units place). Similarly, 123.75 → 124°.

Alternatively, sometimes written as 1.01×10², but for degrees, 101° is fine.

I think it's acceptable.

---

j) tan(θ/3) = 0.3

First, find α = tan⁻¹(0.3) ≈ 16.699° (using calculator)

So θ/3 = 16.699° → θ ≈ 50.097° ≈ 50.1°

Next solution: tan has period 180°, so θ/3 = 16.699° + 180° = 196.699° → θ ≈ 590.097° > 180° ✘

Also, θ/3 must be ≤ 60° (since θ ≤ 180°), so only one solution.

Check: θ/3 ≈ 16.7°, tan(16.7°)≈0.3? tan(16.7)≈0.300 (yes, approximately)

So 50.1°

---

Now, compiling all answers:

a) 15.0°, 75.0°
b) 30.0°, 150°
c) 30.0°, 150°
d) 120°
e) 22.5°, 112.5°
f) 63.4°, 117°
g) 76.7°, 167°
h) 120°
i) 11.3°, 33.8°, 101°, 124°
j) 50.1°

Note: For i, 101.25° to 3 sig fig is 101°, and 123.75° is 124°. Some might argue 101. has three sig fig, but writing 101° is standard.

Double-checking f: cos2θ=-0.6

2θ = arccos(-0.6) ≈ 126.87° and 233.13° → θ=63.435°≈63.4°, 116.565°≈117° — yes.

g: tan2θ=-0.5 → 2θ≈153.435°, 333.435° → θ≈76.717°, 166.717° → 76.7°, 167° — yes.

i: sin4x=√2/2 → 4x=45,135,405,495 → x=11.25,33.75,101.25,123.75 → to 3 sig fig: 11.3, 33.8, 101, 124 — yes.

Final Answer:

a) 15.0°, 75.0°
b) 30.0°, 150°
c) 30.0°, 150°
d) 120°
e) 22.5°, 112.5°
f) 63.4°, 117°
g) 76.7°, 167°
h) 120°
i) 11.3°, 33.8°, 101°, 124°
j) 50.1°