Question Statement

Solve the following trigonometric equations for the values of $\theta$ in the range $0^{\circ }\le \theta \le 360^{\circ }$ (or $0\le \theta \le 2\pi$):

(i) $\cos 5\theta +\cos \theta =\cos 3\theta$ (ii) $\sin \theta =2\sin \left(\frac{\pi }{3}-\theta \right)$ (iii) $\cos 3\theta +2\cos \theta =0$ (iv) $\sin 3\theta =3\sin 2\theta$ (v) $\cos \theta -\cos 3\theta ={\tan }^2\theta$ (vi) $\sin 4\theta +\cos 3\theta =0$ (vii) $2\sin \theta -\tan \theta =6\cot \frac{\theta }{2}$ (viii) $\cos 2\theta +{\csc }^2\theta =0$

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Background and Explanation

To solve these equations, we use trigonometric identities to simplify expressions into a factorable form or a single trigonometric ratio. Key techniques include using sum-to-product formulas, double and triple angle identities, and determining the reference angle to find all solutions within the specified quadrants.

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Solution

Part (i)

$\cos 5\theta +\cos \theta =\cos 3\theta$

Using the sum-to-product formula $\cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$: $2\cos \left(\frac{5\theta +\theta }{2}\right)\cos \left(\frac{5\theta -\theta }{2}\right)=\cos 3\theta$ $2\cos 3\theta \cos 2\theta -\cos 3\theta =0$ $\cos 3\theta (2\cos 2\theta -1)=0$

Case 1: $\cos 3\theta =0$ $3\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\dots$ $\theta =\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}\dots$

Case 2: $2\cos 2\theta -1=0\Rightarrow \cos 2\theta =\frac{1}{2}$ The reference angle is $\frac{\pi }{3}$. Since cosine is positive, $2\theta$ is in the 1st or 4th quadrant.

• 1st Quad: $2\theta =\frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{6}$
• 4th Quad: $2\theta =2\pi -\frac{\pi }{3}=\frac{5\pi }{3}\Rightarrow \theta =\frac{5\pi }{6}$

Solution Set: $S.S=\left\{\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6}\right\}$ (within the first few rotations).

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Part (ii)

$\sin \theta =2\sin \left(\frac{\pi }{3}-\theta \right)$

Using the subtraction formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$: $\sin \theta =2\left[\sin \frac{\pi }{3}\cos \theta -\cos \frac{\pi }{3}\sin \theta \right]$ $\sin \theta =2\left[\frac{\sqrt{3}}{2}\cos \theta -\frac{1}{2}\sin \theta \right]$ $\sin \theta =\sqrt{3}\cos \theta -\sin \theta$ $2\sin \theta =\sqrt{3}\cos \theta$ $\tan \theta =\frac{\sqrt{3}}{2}$

Since $\tan \theta$ is positive, $\theta$ lies in the 1st or 3rd quadrant. Reference angle: $\theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)\approx 40.89^{\circ }$

• 1st Quad: $\theta \approx 40.89^{\circ }\approx \frac{40.89\pi }{180}\text{ rad}$
• 3rd Quad: $\theta =180^{\circ }+40.89^{\circ }=220.89^{\circ }\approx \frac{220.89\pi }{180}\text{ rad}$

Solution Set: $S.S=\left\{\frac{40.89\pi }{180},\frac{220.89\pi }{180}\right\}$

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Part (iii)

$\cos 3\theta +2\cos \theta =0$

Using the triple angle formula $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$: $(4{\cos }^3\theta -3\cos \theta )+2\cos \theta =0$ $4{\cos }^3\theta -\cos \theta =0$ $\cos \theta (4{\cos }^2\theta -1)=0$

Case 1: $\cos \theta =0$ $\theta =\frac{\pi }{2},\frac{3\pi }{2}$

Case 2: $4{\cos }^2\theta -1=0\Rightarrow \cos \theta =\pm \frac{1}{2}$

• If $\cos \theta =\frac{1}{2}$, $\theta =\frac{\pi }{3},\frac{5\pi }{3}$ (Q1, Q4)
• If $\cos \theta =-\frac{1}{2}$, $\theta =\frac{2\pi }{3},\frac{4\pi }{3}$ (Q2, Q3)

Solution Set: $S.S=\left\{\frac{\pi }{3},\frac{\pi }{2},\frac{2\pi }{3},\frac{4\pi }{3},\frac{3\pi }{2},\frac{5\pi }{3}\right\}$

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Part (iv)

$\sin 3\theta =3\sin 2\theta$

Using $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$ and $\sin 2\theta =2\sin \theta \cos \theta$: $3\sin \theta -4{\sin }^3\theta =3(2\sin \theta \cos \theta )$ $3\sin \theta -4{\sin }^3\theta -6\sin \theta \cos \theta =0$ $\sin \theta (3-4{\sin }^2\theta -6\cos \theta )=0$

Case 1: $\sin \theta =0$ $\theta =0,\pi$

Case 2: $3-4(1-{\cos }^2\theta )-6\cos \theta =0$ $3-4+4{\cos }^2\theta -6\cos \theta =0$ $4{\cos }^2\theta -6\cos \theta -1=0$ Using the quadratic formula: $\cos \theta =\frac{6\pm \sqrt{36-4(4)(-1)}}{8}=\frac{6\pm \sqrt{52}}{8}=\frac{3\pm \sqrt{13}}{4}$

• $\cos \theta =\frac{3+\sqrt{13}}{4}\approx 1.65$ (Impossible, as $|\cos \theta |\le 1$)
• $\cos \theta =\frac{3-\sqrt{13}}{4}\approx -0.15$ Reference angle: ${\cos }^{-1}(0.15)\approx 81.37^{\circ }$
  • 2nd Quad: $\theta =180-81.37=98.63^{\circ }$
  • 3rd Quad: $\theta =180+81.37=261.37^{\circ }$

Solution Set: $S.S=\{0,\pi ,98.63^{\circ },261.37^{\circ }\}$

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Part (v)

$\cos \theta -\cos 3\theta ={\tan }^2\theta$

Using $\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)$: $-2\sin (2\theta )\sin (-\theta )=\frac{{\sin }^2\theta }{{\cos }^2\theta }$ $2(2\sin \theta \cos \theta )\sin \theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }$ $4{\sin }^2\theta \cos \theta -\frac{{\sin }^2\theta }{{\cos }^2\theta }=0$ ${\sin }^2\theta \left(\frac{4{\cos }^3\theta -1}{{\cos }^2\theta }\right)=0$

Case 1: ${\sin }^2\theta =0\Rightarrow \theta =0,\pi ,2\pi$ ($0^{\circ },180^{\circ },360^{\circ }$)

Case 2: $4{\cos }^3\theta -1=0\Rightarrow \cos \theta =\sqrt[3]{1/4}\approx 0.63$ Reference angle: ${\cos }^{-1}(0.63)\approx 50.95^{\circ }$

• 1st Quad: $\theta =50.95^{\circ }$
• 4th Quad: $\theta =360^{\circ }-50.95^{\circ }=309.05^{\circ }$

Solution Set: $S.S=\{0^{\circ },50.95^{\circ },180^{\circ },309.05^{\circ },360^{\circ }\}$

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Part (vi)

$\sin 4\theta +\cos 3\theta =0$

Convert $\cos 3\theta$ to $\sin$: $\cos 3\theta =\sin (\frac{\pi }{2}-3\theta )$ $\sin 4\theta +\sin \left(\frac{\pi }{2}-3\theta \right)=0$ Using sum-to-product: $2\sin \left(\frac{4\theta +\frac{\pi }{2}-3\theta }{2}\right)\cos \left(\frac{4\theta -(\frac{\pi }{2}-3\theta )}{2}\right)=0$ $2\sin \left(\frac{\theta +\frac{\pi }{2}}{2}\right)\cos \left(\frac{7\theta -\frac{\pi }{2}}{2}\right)=0$

Case 1: $\sin \left(\frac{\theta +\pi /2}{2}\right)=0$ $\frac{\theta +\pi /2}{2}=0\Rightarrow \theta =-\pi /2$ (Outside range) $\frac{\theta +\pi /2}{2}=\pi \Rightarrow \theta =3\pi /2$

Case 2: $\cos \left(\frac{7\theta -\pi /2}{2}\right)=0$ $\frac{7\theta -\pi /2}{2}=\frac{\pi }{2}\Rightarrow 7\theta =\frac{3\pi }{2}\Rightarrow \theta =\frac{3\pi }{14}$ $\frac{7\theta -\pi /2}{2}=\frac{3\pi }{2}\Rightarrow 7\theta =\frac{7\pi }{2}\Rightarrow \theta =\frac{\pi }{2}$

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Part (vii)

$2\sin \theta -\tan \theta =6\cot \frac{\theta }{2}$

Using $\cot \frac{\theta }{2}=\frac{1+\cos \theta }{\sin \theta }$ and $\tan \theta =\frac{\sin \theta }{\cos \theta }$: $\frac{2\sin \theta \cos \theta -\sin \theta }{\cos \theta }=\frac{6(1+\cos \theta )}{\sin \theta }$ $\sin \theta (2\sin \theta \cos \theta -\sin \theta )=6\cos \theta (1+\cos \theta )$ $2{\sin }^2\theta \cos \theta -{\sin }^2\theta =6\cos \theta +6{\cos }^2\theta$ $2(1-{\cos }^2\theta )\cos \theta -(1-{\cos }^2\theta )=6\cos \theta +6{\cos }^2\theta$ $2{\cos }^3\theta +5{\cos }^2\theta +4\cos \theta +1=0$

By inspection, $\cos \theta =-1$ is a root. Using synthetic division: $2{\cos }^2\theta +3\cos \theta +1=0$ $(2\cos \theta +1)(\cos \theta +1)=0$

• $\cos \theta =-1\Rightarrow \theta =180^{\circ }$
• $\cos \theta =-1/2\Rightarrow \theta =120^{\circ },240^{\circ }$

Solution Set: $S.S=\{120^{\circ },180^{\circ },240^{\circ }\}$

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Part (viii)

$\cos 2\theta +{\csc }^2\theta =0$ $(1-2{\sin }^2\theta )+\frac{1}{{\sin }^2\theta }=0$ ${\sin }^2\theta -2{\sin }^4\theta +1=0$ $2{\sin }^4\theta -{\sin }^2\theta -1=0$ $(2{\sin }^2\theta +1)({\sin }^2\theta -1)=0$

Case 1: ${\sin }^2\theta =1\Rightarrow \sin \theta =\pm 1$ $\theta =90^{\circ },270^{\circ }$

Case 2: ${\sin }^2\theta =-1/2$ (Imaginary, neglect)

Solution Set: $S.S=\{90^{\circ },270^{\circ }\}$

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Key Formulas or Methods Used

• Sum-to-Product: $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$
• Triple Angle: $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$
• Double Angle: $\sin 2\theta =2\sin \theta \cos \theta$
• Half Angle/Identity: $\cot \frac{\theta }{2}=\frac{1+\cos \theta }{\sin \theta }$
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Synthetic Division: Used for solving higher-degree polynomial equations in $\cos \theta$.

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Summary of Steps

1. Simplify: Use identities to convert the equation into a single trig function or a product of factors.
2. Factorize: Set the equation to zero and factor out common terms.
3. Solve Basic Equations: Solve for the trig ratio (e.g., $\cos \theta =k$).
4. Find Reference Angles: Determine the primary angle and use quadrant rules (ASTC) to find all values in $[0,360^{\circ }]$.
5. Check Validity: Ensure values do not make the original equation undefined (e.g., $\tan \theta$ at $90^{\circ }$).