1.05o Trigonometric equations: solve in given intervals

2 Fig. 6 shows the arch ABCD of a bridge. \begin{figure}[h] \end{figure} The section from \(B\) to \(C\) is part of the curve \(O B C E\) with parametric equations $$x = a ( \theta - \sin \theta ) , y = a ( 1 - \cos \theta ) \text { for } 0 \leqslant \theta \leqslant 2 \pi \text {, }$$ where \(a\) is a constant.

1. Find, in terms of \(a\),
   (A) the length of the straight line OE,
   (B) the maximum height of the arch.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). The straight line sections AB and CD are inclined at \(30 ^ { \circ }\) to the horizontal, and are tangents to the curve at B and C respectively. BC is parallel to the \(x\)-axis. BF is parallel to the \(y\)-axis.
3. Show that at the point B the parameter \(\theta\) satisfies the equation $$\sin \theta = \frac { 1 } { \sqrt { 3 } } ( 1 \quad \cos \theta ) .$$ Verify that \(\theta = \frac { 2 } { 3 } \pi\) is a solution of this equation.
   Hence show that \(\mathrm { BF } = \frac { 3 } { 2 } a\), and find OF in terms of \(a\), giving your answer exactly.
4. Find BC and AF in terms of \(a\). Given that the straight line distance AD is 20 metres, calculate the value of \(a\).

2 Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < { } _ { 2 } ^ { 1 } \pi\).-
Hence solve the equation \(4 \cos \theta - \sin \theta = 3\), for \(0 \leqslant \theta \leqslant 2 \pi\).

4 Solve the equation \(\cos 2 \theta = \sin \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\), giving your answers in terms of \(\pi\).

6 Express \(\sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Hence solve the equation \(\sin \theta - 3 \cos \theta = 1\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

1 Express \(6 \cos 2 \theta + \sin \theta\) in terms of \(\sin \theta\).
Hence solve the equation \(6 \cos 2 \theta + \sin \theta = 0\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

2

1. Show that \(\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }\).
2. Hence show that \(\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }\).
3. Hence or otherwise solve the equation \(\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

3 Given the equation \(\sin \left( + 45 ^ { \circ } \right) = 2 \cos [\), show that \(\sin + \cos = 22 \cos\). Hence solve, correct to 2 decimal places, the equation for \(0 ^ { \circ } \quad \left[ \sqrt { 3 } 60 ^ { \circ } \right.\). $$\leqslant \leqslant$$

4 Solve the equation \(\tan \left( \theta + 45 ^ { \circ } \right) = 1 - 2 \tan \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 90 ^ { \circ }\).

5 Given that \(\sin ( \theta + \alpha ) = 2 \sin \theta\), show that \(\tan \theta = \frac { \sin \alpha } { 2 - \cos \alpha }\). Hence solve the equation \(\sin \left( \theta + 40 ^ { \circ } \right) = 2 \sin \theta\), for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

6 Solve the equation \(2 \cos 2 x = 1 + \cos x\), for \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).

2 Express \(3 \cos \theta + 4 \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { - \pi } { 2 }\).
Hence solve the equation \(3 \cos \theta + 4 \sin \theta = 2\) for \(\quad - \pi \leqslant \theta \leqslant \pi\).

3 Show that the equation \(\operatorname { cosec } x + 5 \cot x = 3 \sin x\) may be rearranged as $$3 \cos ^ { 2 } x + 5 \cos x - 2 = 0$$ Hence solve the equation for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\), giving your answers to 1 decimal place.

4 Part of the track of a roller-coaster is modelled by a curve with the parametric equations $$x = 2 \theta - \sin \theta , \quad y = 4 \cos \theta \quad \text { for } 0 \leqslant \theta \leqslant 2 \pi$$ This is shown in Fig. 8. B is a minimum point, and BC is vertical. \begin{figure}[h] \end{figure}

1. Find the values of the parameter at A and B . Hence show that the ratio of the lengths OA and AC is \(( \pi - 1 ) : ( \pi + 1 )\).
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Find the gradient of the track at A .
3. Show that, when the gradient of the track is \(1 , \theta\) satisfies the equation $$\cos \theta - 4 \sin \theta = 2$$
4. Express \(\cos \theta - 4 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\). Hence solve the equation \(\cos \theta - 4 \sin \theta = 2\) for \(0 \leqslant \theta \leqslant 2 \pi\).

5 Express \(4 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\).
Hence solve the equation \(4 \cos \theta - \sin \theta = 3\), for \(0 \leqslant \theta \leqslant 2 \pi\).

2 Show that the equation \(\operatorname { cosec } x + 5 \cot x = 3 \sin x\) may be rearranged as $$3 \cos ^ { 2 } x + 5 \cos x - 2 = 0$$ Hence solve the equation for \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\), giving your answers to 1 decimal place.

5 Solve the equation \(\operatorname { cosec } ^ { 2 } \theta = 1 + 2 \cot \theta\), for \(- 180 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

6 Given that \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\), show that \(\cot ^ { 2 } \theta - \cot \theta - 2 = 0\).
Hence solve the equation \(\operatorname { cosec } ^ { 2 } \theta - \cot \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

8 Solve the equation $$\sec ^ { 2 } \theta = 4 , \quad 0 \leqslant \theta \leqslant \pi ,$$ giving your answers in terms of \(\pi\).

1 Solve the equation \(2 \sec ^ { 2 } \theta = 5 \tan \theta\), for \(0 \leqslant \theta \leqslant \pi\).

2 Solve, correct to 2 decimal places, the equation \(\cot 2 \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

4 Show that \(\cot 2 \theta = \frac { 1 - \tan ^ { 2 } \theta } { 2 \tan \theta }\).
Hence solve the equation $$\cot 2 \theta = 1 + \tan \theta \quad \text { for } 0 ^ { \circ } < \theta < 360 ^ { \circ } .$$

3.Solve for values of \(\theta\) ,in degrees,in the range \(0 \leq \theta \leq 360\) , $$\sqrt { } 2 \times ( \sin 2 \theta + \cos \theta ) + \cos 3 \theta = \sin 2 \theta + \cos \theta$$

3.(a)Use the formulae for \(\sin ( A \pm B )\) and \(\cos ( A \pm B )\) to prove that \(\tan \left( 90 ^ { \circ } - \theta \right) \equiv \cot \theta\) (b)Solve for \(0 < \theta < 360 ^ { \circ }\) $$2 - \sec ^ { 2 } \left( \theta + 11 ^ { \circ } \right) = 2 \tan \left( \theta + 11 ^ { \circ } \right) \tan \left( \theta - 34 ^ { \circ } \right)$$ Give each answer as an integer in degrees.

2.Find the values of \(\tan \theta\) such that $$2 \sin ^ { 2 } \theta - \sin \theta \sec \theta = 2 \sin 2 \theta - 2 .$$

2.Solve,for \(0 < \theta < 2 \pi\) , $$\sin 2 \theta + \cos 2 \theta + 1 = \sqrt { 6 } \cos \theta$$ giving your answers in terms of \(\pi\) .