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Trigonometry, 2ADV T2 2025 HSC 31

The equation  \(\cos \, p x=\dfrac{1}{2}\) has 2 solutions where  \(0 \leq x \leq 2 \pi\)  and  \(p>0\). 

Find all possible values of \(p\).   (3 marks)

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\(\dfrac{5}{6} \leqslant p<\dfrac{7}{6}\)

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\(\cos (p x)=\dfrac{1}{2}\ \ \Rightarrow\ \ px=\cos ^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{3}, \dfrac{5 \pi}{3}, \dfrac{7 \pi}{3}\)

 
\(\text{Since there are 2 solutions in range} \ \ 0 \leq x<2 \pi:\)

\(p x=\dfrac{5 \pi}{3} \ \Rightarrow \ x=\dfrac{5 \pi}{3 p}\)

\(\dfrac{5 \pi}{3 p}\) \(\leqslant 2 \pi\)  
\(3 p\) \(\geqslant \dfrac{5}{2}\)  
\(p\) \(\geqslant \dfrac{5}{6}\)  

 
\(\text{Since}\ \ p x=\dfrac{7 \pi}{3} \ \ \text{cannot be a solution in the range} \ \ 0 \leq x \leq 2 \pi:\) 

\(p x=\dfrac{7 \pi}{3} \ \Rightarrow \ x=\dfrac{7 \pi}{3 p}\)

\(\dfrac{7 \pi}{3 p}\) \(>2 \pi\)  
\(p\) \(<\dfrac{7}{6}\)  

 
\(\therefore \dfrac{5}{6} \leqslant p<\dfrac{7}{6}\)

Trigonometry, 2ADV T2 EQ-Bank 1 MC

Determine the number of values of \(\theta\) in the range  \(0^{\circ} \leqslant \theta \leqslant 360^{\circ}\)  that satisfy the equation

\((\tan \theta-\sqrt{3})(\cos^{2}\theta-1)=0 \)

  1. \(3\)
  2. \(4\)
  3. \(5\)
  4. \(6\)
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\(C\)

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\(\tan \theta = \sqrt{3}\ \rightarrow \text{2 solutions} \)

\(\cos^{2}\theta\) \(=1\)  
\(\cos\theta\) \(= \pm 1\)  
\(\theta\) \(=0^{\circ}, 180^{\circ}, 360^{\circ}\ \rightarrow \text{3 solutions} \)  

 
\(\Rightarrow C\)

Trigonometry, 2ADV T2 2005 HSC 2a

Solve  `cos\ theta =1/sqrt2`  for  `0 ≤ theta ≤ 2pi`.   (2 marks)

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`pi/4, (7pi)/4`

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`cos\ theta = 1/sqrt2,\ \ \ 0 ≤ theta ≤ 2pi`

`text(S)text(ince)\ cos\ pi/4 = 1/sqrt2,\ \ text(and cos)`

`text(is positive in 1st/4th quadrants)`

`theta` `= pi/4, 2pi-pi/4`
  `= pi/4, (7pi)/4`