1.05o Trigonometric equations: solve in given intervals

3 Solve the equation $$\cos \theta + 4 \cos 2 \theta = 3$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

4

1. Show that the equation $$\tan \left( 60 ^ { \circ } + \theta \right) + \tan \left( 60 ^ { \circ } - \theta \right) = k$$ can be written in the form $$( 2 \sqrt { } 3 ) \left( 1 + \tan ^ { 2 } \theta \right) = k \left( 1 - 3 \tan ^ { 2 } \theta \right)$$
2. Hence solve the equation $$\tan \left( 60 ^ { \circ } + \theta \right) + \tan \left( 60 ^ { \circ } - \theta \right) = 3 \sqrt { } 3$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

10

1. It is given that \(2 \tan 2 x + 5 \tan ^ { 2 } x = 0\). Denoting \(\tan x\) by \(t\), form an equation in \(t\) and hence show that either \(t = 0\) or \(t = \sqrt [ 3 ] { } ( t + 0.8 )\).
2. It is given that there is exactly one real value of \(t\) satisfying the equation \(t = \sqrt [ 3 ] { } ( t + 0.8 )\). Verify by calculation that this value lies between 1.2 and 1.3 .
3. Use the iterative formula \(t _ { n + 1 } = \sqrt [ 3 ] { } \left( t _ { n } + 0.8 \right)\) to find the value of \(t\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
4. Using the values of \(t\) found in previous parts of the question, solve the equation $$2 \tan 2 x + 5 \tan ^ { 2 } x = 0$$ for \(- \pi \leqslant x \leqslant \pi\).

2 \includegraphics[max width=\textwidth, alt={}, center]{d3f0b201-3004-497a-9b29-30c94d0bec5b-2_300_767_518_689} In the diagram, \(A B C\) is a triangle in which angle \(A B C\) is a right angle and \(B C = a\). A circular arc, with centre \(C\) and radius \(a\), joins \(B\) and the point \(M\) on \(A C\). The angle \(A C B\) is \(\theta\) radians. The area of the sector \(C M B\) is equal to one third of the area of the triangle \(A B C\).

1. Show that \(\theta\) satisfies the equation $$\tan \theta = 3 \theta .$$
2. This equation has one root in the interval \(0 < \theta < \frac { 1 } { 2 } \pi\). Use the iterative formula $$\theta _ { n + 1 } = \tan ^ { - 1 } \left( 3 \theta _ { n } \right)$$ to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

4 Solve the equation $$\operatorname { cosec } 2 \theta = \sec \theta + \cot \theta$$ giving all solutions in the interval \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

6 It is given that \(\tan 3 x = k \tan x\), where \(k\) is a constant and \(\tan x \neq 0\).

1. By first expanding \(\tan ( 2 x + x )\), show that $$( 3 k - 1 ) \tan ^ { 2 } x = k - 3$$
2. Hence solve the equation \(\tan 3 x = k \tan x\) when \(k = 4\), giving all solutions in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).
3. Show that the equation \(\tan 3 x = k \tan x\) has no root in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\) when \(k = 2\).

3 Solve the equation \(\tan 2 x = 5 \cot x\), for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

3

1. Show that the equation $$\tan \left( x - 60 ^ { \circ } \right) + \cot x = \sqrt { } 3$$ can be written in the form $$2 \tan ^ { 2 } x + ( \sqrt { } 3 ) \tan x - 1 = 0$$
2. Hence solve the equation $$\tan \left( x - 60 ^ { \circ } \right) + \cot x = \sqrt { } 3$$ for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

4

1. Express \(3 \sin \theta + 2 \cos \theta\) in the form \(R \sin ( \theta + \alpha )\), 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 $$3 \sin \theta + 2 \cos \theta = 1$$ for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

3 Solve the equation \(\cot 2 x + \cot x = 3\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

3 By expressing the equation \(\operatorname { cosec } \theta = 3 \sin \theta + \cot \theta\) in terms of \(\cos \theta\) only, solve the equation for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

5

1. Prove the identity \(\cos 4 \theta - 4 \cos 2 \theta \equiv 8 \sin ^ { 4 } \theta - 3\).
2. Hence solve the equation $$\cos 4 \theta = 4 \cos 2 \theta + 3$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

3

1. Express \(( \sqrt { } 5 ) \cos x + 2 \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation $$( \sqrt { } 5 ) \cos \frac { 1 } { 2 } x + 2 \sin \frac { 1 } { 2 } x = 1.2$$ for \(0 ^ { \circ } < x < 360 ^ { \circ }\).

8

1. By first expanding \(2 \sin \left( x - 30 ^ { \circ } \right)\), express \(2 \sin \left( x - 30 ^ { \circ } \right) - \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places. [5]
2. Hence solve the equation $$2 \sin \left( x - 30 ^ { \circ } \right) - \cos x = 1$$ for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

4 By first expressing the equation \(\cot \theta - \cot \left( \theta + 45 ^ { \circ } \right) = 3\) as a quadratic equation in \(\tan \theta\), solve the equation for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

3 Showing all necessary working, solve the equation \(\cot 2 \theta = 2 \tan \theta\) for \(0 ^ { \circ } < \theta < 180 ^ { \circ }\).

2 Express the equation \(\tan \left( \theta + 45 ^ { \circ } \right) - 2 \tan \left( \theta - 45 ^ { \circ } \right) = 4\) as a quadratic equation in \(\tan \theta\). Hence solve this equation for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

4

1. Express \(8 \cos \theta - 15 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), 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 $$8 \cos 2 x - 15 \sin 2 x = 4$$ for \(0 ^ { \circ } < x < 180 ^ { \circ }\).

3

1. Given that \(\sin \left( \theta + 45 ^ { \circ } \right) + 2 \cos \left( \theta + 60 ^ { \circ } \right) = 3 \cos \theta\), find the exact value of \(\tan \theta\) in a form involving surds. You need not simplify your answer.
2. Hence solve the equation \(\sin \left( \theta + 45 ^ { \circ } \right) + 2 \cos \left( \theta + 60 ^ { \circ } \right) = 3 \cos \theta\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).

5

1. Express \(4 \sin \theta - 3 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), stating the value of \(\alpha\) correct to 2 decimal places. Hence
2. solve the equation $$4 \sin \theta - 3 \cos \theta = 2$$ giving all values of \(\theta\) such that \(0 ^ { \circ } < \theta < 360 ^ { \circ }\),
3. write down the greatest value of \(\frac { 1 } { 4 \sin \theta - 3 \cos \theta + 6 }\).

3 Solve the equation $$\cos \theta + 3 \cos 2 \theta = 2$$ giving all solutions in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).

4

1. Show that the equation $$\tan \left( 45 ^ { \circ } + x \right) = 2 \tan \left( 45 ^ { \circ } - x \right)$$ can be written in the form $$\tan ^ { 2 } x - 6 \tan x + 1 = 0$$
2. Hence solve the equation \(\tan \left( 45 ^ { \circ } + x \right) = 2 \tan \left( 45 ^ { \circ } - x \right)\), for \(0 ^ { \circ } < x < 90 ^ { \circ }\).

5 By expressing \(8 \sin \theta - 6 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), solve the equation $$8 \sin \theta - 6 \cos \theta = 7$$ for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).

2 Solve the equation $$\tan x \tan 2 x = 1 ,$$ giving all solutions in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).

4 The curve with equation \(y = \mathrm { e } ^ { - x } \sin x\) has one stationary point for which \(0 \leqslant x \leqslant \pi\).

1. Find the \(x\)-coordinate of this point.
2. Determine whether this point is a maximum or a minimum point.