Eleven equations, each solved from scratch by factoring into simple ratio-equations (sum-to-product for a sum/difference of sines or cosines, a Pythagorean substitution to reach one ratio, the a cos θ + b sin θ a\cos\theta+b\sin\theta a cos θ + b sin θ
Part (i). sin 5 x − sin x = cos 3 x \sin5x-\sin x=\cos3x sin 5 x − sin x = cos 3 x
Step 1. Sum-to-product: sin 5 x − sin x = 2 cos ( 5 x + x 2 ) sin ( 5 x − x 2 ) = 2 cos 3 x sin 2 x \sin5x-\sin x=2\cos\!\left(\dfrac{5x+x}2\right)\sin\!\left(\dfrac{5x-x}2\right)=2\cos3x\sin2x sin 5 x − sin x = 2 cos ( 2 5 x + x ) sin ( 2 5 x − x ) = 2 cos 3 x sin 2 x
Step 2. So 2 cos 3 x sin 2 x = cos 3 x ⇒ cos 3 x ( 2 sin 2 x − 1 ) = 0 2\cos3x\sin2x=\cos3x \Rightarrow \cos3x(2\sin2x-1)=0 2 cos 3 x sin 2 x = cos 3 x ⇒ cos 3 x ( 2 sin 2 x − 1 ) = 0
Step 3. Either cos 3 x = 0 ⇒ 3 x = ( 2 n + 1 ) π 2 ⇒ x = ( 2 n + 1 ) π 6 , n ∈ Z \cos3x=0 \Rightarrow 3x=(2n+1)\dfrac\pi2 \Rightarrow x=(2n+1)\dfrac\pi6,\ n\in\mathbb Z cos 3 x = 0 ⇒ 3 x = ( 2 n + 1 ) 2 π ⇒ x = ( 2 n + 1 ) 6 π , n ∈ Z sin 2 x = 1 2 = sin π 6 ⇒ 2 x = n π + ( − 1 ) n π 6 ⇒ x = n π 2 + ( − 1 ) n π 12 , n ∈ Z \sin2x=\dfrac12=\sin\dfrac\pi6 \Rightarrow 2x=n\pi+(-1)^n\dfrac\pi6 \Rightarrow x=\dfrac{n\pi}2+(-1)^n\dfrac\pi{12},\ n\in\mathbb Z sin 2 x = 2 1 = sin 6 π ⇒ 2 x = nπ + ( − 1 ) n 6 π ⇒ x = 2 nπ + ( − 1 ) n 12 π , n ∈ Z
Step 4 (check). n = 0 n=0 n = 0 x = π 12 = 15 ∘ x=\dfrac\pi{12}=15^\circ x = 12 π = 1 5 ∘ sin 75 ∘ ≈ 0.9659 \sin75^\circ\approx0.9659 sin 7 5 ∘ ≈ 0.9659 sin 15 ∘ ≈ 0.2588 \sin15^\circ\approx0.2588 sin 1 5 ∘ ≈ 0.2588 ≈ 0.7071 \approx0.7071 ≈ 0.7071 cos 45 ∘ ≈ 0.7071 \cos45^\circ\approx0.7071 cos 4 5 ∘ ≈ 0.7071 ✓ .
Part (ii). 2 cos 2 θ + 3 sin θ − 3 = 0 2\cos^2\theta+3\sin\theta-3=0 2 cos 2 θ + 3 sin θ − 3 = 0
Step 1. Replace cos 2 θ = 1 − sin 2 θ \cos^2\theta=1-\sin^2\theta cos 2 θ = 1 − sin 2 θ 2 ( 1 − sin 2 θ ) + 3 sin θ − 3 = 0 ⇒ − 2 sin 2 θ + 3 sin θ − 1 = 0 ⇒ 2 sin 2 θ − 3 sin θ + 1 = 0 2(1-\sin^2\theta)+3\sin\theta-3=0 \Rightarrow -2\sin^2\theta+3\sin\theta-1=0 \Rightarrow 2\sin^2\theta-3\sin\theta+1=0 2 ( 1 − sin 2 θ ) + 3 sin θ − 3 = 0 ⇒ − 2 sin 2 θ + 3 sin θ − 1 = 0 ⇒ 2 sin 2 θ − 3 sin θ + 1 = 0
Step 2. Factor: ( 2 sin θ − 1 ) ( sin θ − 1 ) = 0 (2\sin\theta-1)(\sin\theta-1)=0 ( 2 sin θ − 1 ) ( sin θ − 1 ) = 0 sin θ = 1 2 \sin\theta=\dfrac12 sin θ = 2 1 sin θ = 1 \sin\theta=1 sin θ = 1
Step 3. sin θ = 1 2 = sin π 6 ⇒ θ = n π + ( − 1 ) n π 6 \sin\theta=\dfrac12=\sin\dfrac\pi6 \Rightarrow \theta=n\pi+(-1)^n\dfrac\pi6 sin θ = 2 1 = sin 6 π ⇒ θ = nπ + ( − 1 ) n 6 π sin θ = 1 = sin π 2 ⇒ θ = n π + ( − 1 ) n π 2 \sin\theta=1=\sin\dfrac\pi2 \Rightarrow \theta=n\pi+(-1)^n\dfrac\pi2 sin θ = 1 = sin 2 π ⇒ θ = nπ + ( − 1 ) n 2 π
Step 4 (check). sin θ = 1 2 \sin\theta=\dfrac12 sin θ = 2 1 2 cos 2 θ = 2 ( 1 − 1 4 ) = 1.5 2\cos^2\theta=2\left(1-\tfrac14\right)=1.5 2 cos 2 θ = 2 ( 1 − 4 1 ) = 1.5 3 ( 1 2 ) = 1.5 3\left(\tfrac12\right)=1.5 3 ( 2 1 ) = 1.5 3 3 3 2 cos 2 θ + 3 sin θ = 3 2\cos^2\theta+3\sin\theta=3 2 cos 2 θ + 3 sin θ = 3 ✓ .
Part (iii). cos θ + cos 3 θ = 2 cos 2 θ \cos\theta+\cos3\theta=2\cos2\theta cos θ + cos 3 θ = 2 cos 2 θ
Step 1. Sum-to-product: cos θ + cos 3 θ = 2 cos ( θ + 3 θ 2 ) cos ( θ − 3 θ 2 ) = 2 cos 2 θ cos θ \cos\theta+\cos3\theta=2\cos\!\left(\dfrac{\theta+3\theta}2\right)\cos\!\left(\dfrac{\theta-3\theta}2\right)=2\cos2\theta\cos\theta cos θ + cos 3 θ = 2 cos ( 2 θ + 3 θ ) cos ( 2 θ − 3 θ ) = 2 cos 2 θ cos θ cos ( − θ ) = cos θ \cos(-\theta)=\cos\theta cos ( − θ ) = cos θ
Step 2. So 2 cos 2 θ cos θ = 2 cos 2 θ ⇒ 2 cos 2 θ ( cos θ − 1 ) = 0 2\cos2\theta\cos\theta=2\cos2\theta \Rightarrow 2\cos2\theta(\cos\theta-1)=0 2 cos 2 θ cos θ = 2 cos 2 θ ⇒ 2 cos 2 θ ( cos θ − 1 ) = 0
Step 3. Either cos 2 θ = 0 ⇒ 2 θ = ( 2 n + 1 ) π 2 ⇒ θ = ( 2 n + 1 ) π 4 , n ∈ Z \cos2\theta=0 \Rightarrow 2\theta=(2n+1)\dfrac\pi2 \Rightarrow \theta=(2n+1)\dfrac\pi4,\ n\in\mathbb Z cos 2 θ = 0 ⇒ 2 θ = ( 2 n + 1 ) 2 π ⇒ θ = ( 2 n + 1 ) 4 π , n ∈ Z cos θ = 1 ⇒ θ = 2 n π , n ∈ Z \cos\theta=1 \Rightarrow \theta=2n\pi,\ n\in\mathbb Z cos θ = 1 ⇒ θ = 2 nπ , n ∈ Z
Step 4 (check). θ = π 4 \theta=\dfrac\pi4 θ = 4 π cos 45 ∘ + cos 135 ∘ = 0.7071 − 0.7071 = 0 \cos45^\circ+\cos135^\circ=0.7071-0.7071=0 cos 4 5 ∘ + cos 13 5 ∘ = 0.7071 − 0.7071 = 0 2 cos 90 ∘ = 0 2\cos90^\circ=0 2 cos 9 0 ∘ = 0 ✓ .
Part (iv). sin θ + sin 3 θ + sin 5 θ = 0 \sin\theta+\sin3\theta+\sin5\theta=0 sin θ + sin 3 θ + sin 5 θ = 0
Step 1. Pair the outer terms: sin θ + sin 5 θ = 2 sin ( θ + 5 θ 2 ) cos ( θ − 5 θ 2 ) = 2 sin 3 θ cos 2 θ \sin\theta+\sin5\theta=2\sin\!\left(\dfrac{\theta+5\theta}2\right)\cos\!\left(\dfrac{\theta-5\theta}2\right)=2\sin3\theta\cos2\theta sin θ + sin 5 θ = 2 sin ( 2 θ + 5 θ ) cos ( 2 θ − 5 θ ) = 2 sin 3 θ cos 2 θ
Step 2. So the equation becomes 2 sin 3 θ cos 2 θ + sin 3 θ = 0 ⇒ sin 3 θ ( 2 cos 2 θ + 1 ) = 0 2\sin3\theta\cos2\theta+\sin3\theta=0 \Rightarrow \sin3\theta(2\cos2\theta+1)=0 2 sin 3 θ cos 2 θ + sin 3 θ = 0 ⇒ sin 3 θ ( 2 cos 2 θ + 1 ) = 0
Step 3. Either sin 3 θ = 0 ⇒ 3 θ = n π ⇒ θ = n π 3 , n ∈ Z \sin3\theta=0 \Rightarrow 3\theta=n\pi \Rightarrow \theta=\dfrac{n\pi}3,\ n\in\mathbb Z sin 3 θ = 0 ⇒ 3 θ = nπ ⇒ θ = 3 nπ , n ∈ Z cos 2 θ = − 1 2 = cos 2 π 3 ⇒ 2 θ = 2 n π ± 2 π 3 ⇒ θ = n π ± π 3 , n ∈ Z \cos2\theta=-\dfrac12=\cos\dfrac{2\pi}3 \Rightarrow 2\theta=2n\pi\pm\dfrac{2\pi}3 \Rightarrow \theta=n\pi\pm\dfrac\pi3,\ n\in\mathbb Z cos 2 θ = − 2 1 = cos 3 2 π ⇒ 2 θ = 2 nπ ± 3 2 π ⇒ θ = nπ ± 3 π , n ∈ Z
Step 4 (check). θ = π 3 \theta=\dfrac\pi3 θ = 3 π sin 60 ∘ + sin 180 ∘ + sin 300 ∘ = 0.866 + 0 − 0.866 = 0 \sin60^\circ+\sin180^\circ+\sin300^\circ=0.866+0-0.866=0 sin 6 0 ∘ + sin 18 0 ∘ + sin 30 0 ∘ = 0.866 + 0 − 0.866 = 0 ✓ . (Every value produced by the second family is in fact already a multiple of π 3 \dfrac\pi3 3 π
Part (v). sin 2 θ − cos 2 θ − sin θ + cos θ = 0 \sin2\theta-\cos2\theta-\sin\theta+\cos\theta=0 sin 2 θ − cos 2 θ − sin θ + cos θ = 0
Step 1. Group and apply sum-to-product to each pair: sin 2 θ − sin θ = 2 cos ( 3 θ 2 ) sin ( θ 2 ) \sin2\theta-\sin\theta=2\cos\!\left(\dfrac{3\theta}2\right)\sin\!\left(\dfrac\theta2\right) sin 2 θ − sin θ = 2 cos ( 2 3 θ ) sin ( 2 θ ) cos 2 θ − cos θ = − 2 sin ( 3 θ 2 ) sin ( θ 2 ) \cos2\theta-\cos\theta=-2\sin\!\left(\dfrac{3\theta}2\right)\sin\!\left(\dfrac\theta2\right) cos 2 θ − cos θ = − 2 sin ( 2 3 θ ) sin ( 2 θ )
Step 2. So ( sin 2 θ − sin θ ) − ( cos 2 θ − cos θ ) = 2 cos ( 3 θ 2 ) sin ( θ 2 ) + 2 sin ( 3 θ 2 ) sin ( θ 2 ) = 2 sin ( θ 2 ) [ cos ( 3 θ 2 ) + sin ( 3 θ 2 ) ] = 0 (\sin2\theta-\sin\theta)-(\cos2\theta-\cos\theta)=2\cos\!\left(\tfrac{3\theta}2\right)\sin\!\left(\tfrac\theta2\right)+2\sin\!\left(\tfrac{3\theta}2\right)\sin\!\left(\tfrac\theta2\right)=2\sin\!\left(\tfrac\theta2\right)\left[\cos\!\left(\tfrac{3\theta}2\right)+\sin\!\left(\tfrac{3\theta}2\right)\right]=0 ( sin 2 θ − sin θ ) − ( cos 2 θ − cos θ ) = 2 cos ( 2 3 θ ) sin ( 2 θ ) + 2 sin ( 2 3 θ ) sin ( 2 θ ) = 2 sin ( 2 θ ) [ cos ( 2 3 θ ) + sin ( 2 3 θ ) ] = 0
Step 3. Either sin θ 2 = 0 ⇒ θ 2 = n π ⇒ θ = 2 n π , n ∈ Z \sin\dfrac\theta2=0 \Rightarrow \dfrac\theta2=n\pi \Rightarrow \theta=2n\pi,\ n\in\mathbb Z sin 2 θ = 0 ⇒ 2 θ = nπ ⇒ θ = 2 nπ , n ∈ Z cos 3 θ 2 + sin 3 θ 2 = 0 ⇒ tan 3 θ 2 = − 1 = tan ( − π 4 ) ⇒ 3 θ 2 = n π − π 4 ⇒ θ = 2 n π 3 − π 6 , n ∈ Z \cos\dfrac{3\theta}2+\sin\dfrac{3\theta}2=0 \Rightarrow \tan\dfrac{3\theta}2=-1=\tan\!\left(-\dfrac\pi4\right) \Rightarrow \dfrac{3\theta}2=n\pi-\dfrac\pi4 \Rightarrow \theta=\dfrac{2n\pi}3-\dfrac\pi6,\ n\in\mathbb Z cos 2 3 θ + sin 2 3 θ = 0 ⇒ tan 2 3 θ = − 1 = tan ( − 4 π ) ⇒ 2 3 θ = nπ − 4 π ⇒ θ = 3 2 nπ − 6 π , n ∈ Z
Step 4 (check). n = 1 n=1 n = 1 θ = 2 π 3 − π 6 = π 2 \theta=\dfrac{2\pi}3-\dfrac\pi6=\dfrac\pi2 θ = 3 2 π − 6 π = 2 π sin π − cos π − sin π 2 + cos π 2 = 0 − ( − 1 ) − 1 + 0 = 0 \sin\pi-\cos\pi-\sin\dfrac\pi2+\cos\dfrac\pi2=0-(-1)-1+0=0 sin π − cos π − sin 2 π + cos 2 π = 0 − ( − 1 ) − 1 + 0 = 0 ✓ . θ = 0 \theta=0 θ = 0 n = 0 n=0 n = 0 0 − 1 − 0 + 1 = 0 0-1-0+1=0 0 − 1 − 0 + 1 = 0 ✓ .
Part (vi). sin θ + cos θ = 2 \sin\theta+\cos\theta=\sqrt2 sin θ + cos θ = 2
Step 1. Divide by r = 1 2 + 1 2 = 2 r=\sqrt{1^2+1^2}=\sqrt2 r = 1 2 + 1 2 = 2 1 2 sin θ + 1 2 cos θ = 1 \dfrac1{\sqrt2}\sin\theta+\dfrac1{\sqrt2}\cos\theta=1 2 1 sin θ + 2 1 cos θ = 1
Step 2. Recognise 1 2 = sin π 4 = cos π 4 \dfrac1{\sqrt2}=\sin\dfrac\pi4=\cos\dfrac\pi4 2 1 = sin 4 π = cos 4 π sin θ sin π 4 + cos θ cos π 4 = cos ( θ − π 4 ) \sin\theta\sin\dfrac\pi4+\cos\theta\cos\dfrac\pi4=\cos\!\left(\theta-\dfrac\pi4\right) sin θ sin 4 π + cos θ cos 4 π = cos ( θ − 4 π ) cos A cos B + sin A sin B = cos ( A − B ) \cos A\cos B+\sin A\sin B=\cos(A-B) cos A cos B + sin A sin B = cos ( A − B )
Step 3. So cos ( θ − π 4 ) = 1 ⇒ θ − π 4 = 2 n π ⇒ θ = 2 n π + π 4 , n ∈ Z \cos\!\left(\theta-\dfrac\pi4\right)=1 \Rightarrow \theta-\dfrac\pi4=2n\pi \Rightarrow \theta=2n\pi+\dfrac\pi4,\ n\in\mathbb Z cos ( θ − 4 π ) = 1 ⇒ θ − 4 π = 2 nπ ⇒ θ = 2 nπ + 4 π , n ∈ Z
Step 4 (check). θ = π 4 \theta=\dfrac\pi4 θ = 4 π sin 45 ∘ + cos 45 ∘ = 1 2 + 1 2 = 2 \sin45^\circ+\cos45^\circ=\dfrac1{\sqrt2}+\dfrac1{\sqrt2}=\sqrt2 sin 4 5 ∘ + cos 4 5 ∘ = 2 1 + 2 1 = 2 ✓ .
Part (vii). sin θ + 3 cos θ = 1 \sin\theta+\sqrt3\cos\theta=1 sin θ + 3 cos θ = 1
Step 1. Divide by r = 1 2 + ( 3 ) 2 = 2 r=\sqrt{1^2+(\sqrt3)^2}=2 r = 1 2 + ( 3 ) 2 = 2 1 2 sin θ + 3 2 cos θ = 1 2 \dfrac12\sin\theta+\dfrac{\sqrt3}2\cos\theta=\dfrac12 2 1 sin θ + 2 3 cos θ = 2 1
Step 2. Recognise 1 2 = sin π 6 \dfrac12=\sin\dfrac\pi6 2 1 = sin 6 π 3 2 = cos π 6 \dfrac{\sqrt3}2=\cos\dfrac\pi6 2 3 = cos 6 π sin θ sin π 6 + cos θ cos π 6 = cos ( θ − π 6 ) \sin\theta\sin\dfrac\pi6+\cos\theta\cos\dfrac\pi6=\cos\!\left(\theta-\dfrac\pi6\right) sin θ sin 6 π + cos θ cos 6 π = cos ( θ − 6 π )
Step 3. So cos ( θ − π 6 ) = 1 2 = cos π 3 ⇒ θ − π 6 = 2 n π ± π 3 ⇒ θ = 2 n π + π 6 + π 3 = 2 n π + π 2 , or θ = 2 n π + π 6 − π 3 = 2 n π − π 6 , n ∈ Z \cos\!\left(\theta-\dfrac\pi6\right)=\dfrac12=\cos\dfrac\pi3 \Rightarrow \theta-\dfrac\pi6=2n\pi\pm\dfrac\pi3 \Rightarrow \theta=2n\pi+\dfrac\pi6+\dfrac\pi3=2n\pi+\dfrac\pi2,\ \text{ or }\ \theta=2n\pi+\dfrac\pi6-\dfrac\pi3=2n\pi-\dfrac\pi6,\ n\in\mathbb Z cos ( θ − 6 π ) = 2 1 = cos 3 π ⇒ θ − 6 π = 2 nπ ± 3 π ⇒ θ = 2 nπ + 6 π + 3 π = 2 nπ + 2 π , or θ = 2 nπ + 6 π − 3 π = 2 nπ − 6 π , n ∈ Z
Step 4 (check). θ = π 2 \theta=\dfrac\pi2 θ = 2 π sin 90 ∘ + 3 cos 90 ∘ = 1 + 0 = 1 \sin90^\circ+\sqrt3\cos90^\circ=1+0=1 sin 9 0 ∘ + 3 cos 9 0 ∘ = 1 + 0 = 1 ✓ . θ = − π 6 \theta=-\dfrac\pi6 θ = − 6 π sin ( − 30 ∘ ) + 3 cos ( − 30 ∘ ) = − 1 2 + 3 ⋅ 3 2 = − 1 2 + 3 2 = 1 \sin(-30^\circ)+\sqrt3\cos(-30^\circ)=-\dfrac12+\sqrt3\cdot\dfrac{\sqrt3}2=-\dfrac12+\dfrac32=1 sin ( − 3 0 ∘ ) + 3 cos ( − 3 0 ∘ ) = − 2 1 + 3 ⋅ 2 3 = − 2 1 + 2 3 = 1 ✓ .
Part (viii). cot θ + cosec θ = 3 \cot\theta+\operatorname{cosec}\theta=\sqrt3 cot θ + cosec θ = 3
Step 1. Write both in terms of sine/cosine: cos θ sin θ + 1 sin θ = cos θ + 1 sin θ = 3 \dfrac{\cos\theta}{\sin\theta}+\dfrac1{\sin\theta}=\dfrac{\cos\theta+1}{\sin\theta}=\sqrt3 sin θ cos θ + sin θ 1 = sin θ cos θ + 1 = 3
Step 2. Use the half-angle identities cos θ + 1 = 2 cos 2 θ 2 \cos\theta+1=2\cos^2\dfrac\theta2 cos θ + 1 = 2 cos 2 2 θ sin θ = 2 sin θ 2 cos θ 2 \sin\theta=2\sin\dfrac\theta2\cos\dfrac\theta2 sin θ = 2 sin 2 θ cos 2 θ 2 cos 2 θ 2 2 sin θ 2 cos θ 2 = cot θ 2 = 3 = cot π 6 \dfrac{2\cos^2\frac\theta2}{2\sin\frac\theta2\cos\frac\theta2}=\cot\dfrac\theta2=\sqrt3=\cot\dfrac\pi6 2 sin 2 θ cos 2 θ 2 cos 2 2 θ = cot 2 θ = 3 = cot 6 π
Step 3. So cot θ 2 = cot π 6 ⇒ θ 2 = n π + π 6 ⇒ θ = 2 n π + π 3 , n ∈ Z \cot\dfrac\theta2=\cot\dfrac\pi6 \Rightarrow \dfrac\theta2=n\pi+\dfrac\pi6 \Rightarrow \theta=2n\pi+\dfrac\pi3,\ n\in\mathbb Z cot 2 θ = cot 6 π ⇒ 2 θ = nπ + 6 π ⇒ θ = 2 nπ + 3 π , n ∈ Z
Step 4 (domain check). cot θ \cot\theta cot θ cosec θ \operatorname{cosec}\theta cosec θ sin θ ≠ 0 \sin\theta\ne0 sin θ = 0 θ = 2 n π + π 3 \theta=2n\pi+\dfrac\pi3 θ = 2 nπ + 3 π π \pi π cos θ + 1 = 3 sin θ \cos\theta+1=\sqrt3\sin\theta cos θ + 1 = 3 sin θ θ = ( 2 n + 1 ) π \theta=(2n+1)\pi θ = ( 2 n + 1 ) π sin θ = 0 \sin\theta=0 sin θ = 0 cot θ , cosec θ \cot\theta,\operatorname{cosec}\theta cot θ , cosec θ
Step 5 (numeric check). θ = π 3 = 60 ∘ \theta=\dfrac\pi3=60^\circ θ = 3 π = 6 0 ∘ cot 60 ∘ + cosec 60 ∘ = 1 3 + 2 3 = 3 3 = 3 \cot60^\circ+\operatorname{cosec}60^\circ=\dfrac1{\sqrt3}+\dfrac2{\sqrt3}=\dfrac3{\sqrt3}=\sqrt3 cot 6 0 ∘ + cosec 6 0 ∘ = 3 1 + 3 2 = 3 3 = 3 ✓ . …