1.05o Trigonometric equations: solve in given intervals

4

1. Show that the equation $$\sin \left( x + 30 ^ { \circ } \right) = 2 \cos \left( x + 60 ^ { \circ } \right)$$ can be written in the form $$( 3 \sqrt { } 3 ) \sin x = \cos x$$
2. Hence solve the equation $$\sin \left( x + 30 ^ { \circ } \right) = 2 \cos \left( x + 60 ^ { \circ } \right)$$ for \(- 180 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).

4

1. Show that the equation \(\sin \left( 60 ^ { \circ } - x \right) = 2 \sin x\) can be written in the form \(\tan x = k\), where \(k\) is a constant.
2. Hence solve the equation \(\sin \left( 60 ^ { \circ } - x \right) = 2 \sin x\), for \(0 ^ { \circ } < x < 360 ^ { \circ }\).

6

1. Express \(3 \cos x + 4 \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), stating the exact value of \(R\) and giving the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation $$3 \cos x + 4 \sin x = 4.5$$ giving all solutions in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\).

5 Solve the equation \(8 + \cot \theta = 2 \operatorname { cosec } ^ { 2 } \theta\), giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

6

1. Express \(2 \sin \theta - \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation $$2 \sin \theta - \cos \theta = - 0.4$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

8

1. Express \(5 \cos \theta - 3 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation $$5 \cos \theta - 3 \sin \theta = 4$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
3. Write down the least value of \(15 \cos \theta - 9 \sin \theta\) as \(\theta\) varies.

5 Solve the equation \(5 \sec ^ { 2 } 2 \theta = \tan 2 \theta + 9\), giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

3 Solve the equation $$2 \cos 2 \theta = 4 \cos \theta - 3$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

8

1. Given that \(\tan A = t\) and \(\tan ( A + B ) = 4\), find \(\tan B\) in terms of \(t\).
2. Solve the equation $$2 \tan \left( 45 ^ { \circ } - x \right) = 3 \tan x$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

6

1. Find
   1. \(\int \frac { \mathrm { e } ^ { 2 x } + 6 } { \mathrm { e } ^ { 2 x } } \mathrm {~d} x\),
   2. \(\int 3 \cos ^ { 2 } x \mathrm {~d} x\).
2. Use the trapezium rule with 2 intervals to estimate the value of $$\int _ { 1 } ^ { 2 } \frac { 6 } { \ln ( x + 2 ) } \mathrm { d } x$$ giving your answer correct to 2 decimal places.
   1. Express \(3 \cos \theta + \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
   2. Hence solve the equation $$3 \cos 2 x + \sin 2 x = 2$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).

3 Solve the equation \(2 \cot ^ { 2 } \theta - 5 \operatorname { cosec } \theta = 10\), giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

7

1. Express \(5 \cos \theta - 12 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation \(5 \cos \theta - 12 \sin \theta = 8\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
3. Find the greatest possible value of $$7 + 5 \cos \frac { 1 } { 2 } \phi - 12 \sin \frac { 1 } { 2 } \phi$$ as \(\phi\) varies, and determine the smallest positive value of \(\phi\) for which this greatest value occurs.
   [0pt] [4]

3

1. Express \(8 \sin \theta + 15 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation $$8 \sin \theta + 15 \cos \theta = 6$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

4 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = 6 x ^ { 3 } + 11 x ^ { 2 } + a x + a$$ where \(a\) is a constant. It is given that \(( x + 2 )\) is a factor of \(\mathrm { p } ( x )\).

1. Use the factor theorem to show that \(a = - 4\).
2. When \(a = - 4\),
   1. factorise \(\mathrm { p } ( x )\) completely,
   2. solve the equation \(6 \sec ^ { 3 } \theta + 11 \sec ^ { 2 } \theta + a \sec \theta + a = 0\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

6

1. Express \(( \sqrt { } 5 ) \cos \theta - 2 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the value of \(\alpha\) correct to 2 decimal places.
2. Hence
   1. solve the equation \(( \sqrt { } 5 ) \cos \theta - 2 \sin \theta = 0.9\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\),
   2. state the greatest and least values of $$10 + ( \sqrt { } 5 ) \cos \theta - 2 \sin \theta$$ as \(\theta\) varies.

2 Solve the equation \(5 \cos \theta ( 1 + \cos 2 \theta ) = 4\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

3 Solve the equation \(\sec ^ { 2 } \theta = 3 \operatorname { cosec } \theta\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

7 \includegraphics[max width=\textwidth, alt={}, center]{6bf7ba66-8362-4ac0-8e5c-3f88a3ccdf86-12_424_488_260_826} The diagram shows the curve with equation \(y = \sin 2 x + 3 \cos 2 x\) for \(0 \leqslant x \leqslant \pi\). At the points \(P\) and \(Q\) on the curve, the gradient of the curve is 3 .

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. By first expressing \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in the form \(R \cos ( 2 x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), find the \(x\)-coordinates of \(P\) and \(Q\), giving your answers correct to 4 significant figures.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

3 \includegraphics[max width=\textwidth, alt={}, center]{cc7e798e-0817-405c-bae0-b24b9f451fbf-04_378_486_260_826} The diagram shows the curve with equation $$y = 5 \sin 2 x - 3 \tan 2 x$$ for values of \(x\) such that \(0 \leqslant x < \frac { 1 } { 4 } \pi\). Find the \(x\)-coordinate of the stationary point \(M\), giving your answer correct to 3 significant figures.

7

1. Use the factor theorem to show that ( \(2 x + 3\) ) is a factor of $$8 x ^ { 3 } + 4 x ^ { 2 } - 10 x + 3$$
2. Show that the equation \(2 \cos 2 \theta = \frac { 6 \cos \theta - 5 } { 2 \cos \theta + 1 }\) can be expressed as $$8 \cos ^ { 3 } \theta + 4 \cos ^ { 2 } \theta - 10 \cos \theta + 3 = 0 .$$
3. Solve the equation \(2 \cos 2 \theta = \frac { 6 \cos \theta - 5 } { 2 \cos \theta + 1 }\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

6

1. Showing all necessary working, solve the equation $$\sec \alpha \operatorname { cosec } \alpha = 7$$ for \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\).
2. Showing all necessary working, solve the equation $$\sin \left( \beta + 20 ^ { \circ } \right) + \sin \left( \beta - 20 ^ { \circ } \right) = 6 \cos \beta$$ for \(0 ^ { \circ } < \beta < 90 ^ { \circ }\).

3

1. Express \(8 \sin \theta + 15 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation $$8 \sin \theta + 15 \cos \theta = 6$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

3 Express the equation \(\tan \left( \theta + 60 ^ { \circ } \right) = 2 + \tan \left( 60 ^ { \circ } - \theta \right)\) as a quadratic equation in \(\tan \theta\), and hence solve the equation for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

4 A curve has equation \(y = \cos x \sin 2 x\).
Find the \(x\)-coordinate of the stationary point in the interval \(0 < x < \frac { 1 } { 2 } \pi\), giving your answer correct to 3 significant figures.

5

1. Express \(\sqrt { 2 } \cos x - \sqrt { 5 } \sin x\) in the form \(R \cos ( x + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\) and the value of \(\alpha\) correct to 3 decimal places.
2. Hence solve the equation \(\sqrt { 2 } \cos 2 \theta - \sqrt { 5 } \sin 2 \theta = 1\), for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).