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1.05o
1.05o
Trigonometric equations: solve in given intervals
1022 questions
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Edexcel C3 Q1
6 marks
Standard +0.8
Solve the equation
$$3 \operatorname { cosec } \theta ^ { \circ } + 8 \cos \theta ^ { \circ } = 0$$ for \(\theta\) in the interval \(0 \leq \theta \leq 180\), giving your answers to 1 decimal place.
Edexcel C3 Q2
9 marks
Standard +0.3
2. (a) Use the identities for \(\cos ( A + B )\) and \(\cos ( A - B )\) to prove that $$2 \cos A \cos B \equiv \cos ( A + B ) + \cos ( A - B ) .$$ (b) Hence, or otherwise, find in terms of \(\pi\) the solutions of the equation $$2 \cos \left( x + \frac { \pi } { 2 } \right) = \sec \left( x + \frac { \pi } { 6 } \right) ,$$ for \(x\) in the interval \(0 \leq x \leq \pi\).
Edexcel C3 Q4
10 marks
Standard +0.3
4. (a) Express \(2 \sin x ^ { \circ } - 3 \cos x ^ { \circ }\) in the form \(R \sin ( x - \alpha ) ^ { \circ }\) where \(R > 0\) and \(0 < \alpha < 90\).
(b) Show that the equation $$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2$$ can be written in the form $$2 \sin x ^ { \circ } - 3 \cos x ^ { \circ } = 1 .$$ (c) Solve the equation $$\operatorname { cosec } x ^ { \circ } + 3 \cot x ^ { \circ } = 2 ,$$ for \(x\) in the interval \(0 \leq x \leq 360\), giving your answers to 1 decimal place.
Edexcel C3 Q2
7 marks
Challenging +1.2
2. Giving your answers to 1 decimal place, solve the equation $$5 \tan ^ { 2 } 2 \theta - 13 \sec 2 \theta = 1 ,$$ for \(\theta\) in the interval \(0 \leq \theta \leq 360 ^ { \circ }\).
Edexcel C3 Q6
11 marks
Standard +0.8
(a) Use the identities for \(\cos ( A + B )\) and \(\cos ( A - B )\) to prove that
$$\cos P - \cos Q \equiv - 2 \sin \frac { P + Q } { 2 } \sin \frac { P - Q } { 2 }$$ (b) Hence find all solutions in the interval \(0 \leq x < 180\) to the equation $$\cos 5 x ^ { \circ } + \sin 3 x ^ { \circ } - \cos x ^ { \circ } = 0$$
Edexcel C3 Q6
11 marks
Standard +0.8
6. (a) Prove the identity $$2 \cot 2 x + \tan x \equiv \cot x , \quad x \neq \frac { n } { 2 } \pi , \quad n \in \mathbb { Z } .$$ (b) Solve, for \(0 \leq x < \pi\), the equation $$2 \cot 2 x + \tan x = \operatorname { cosec } ^ { 2 } x - 7 ,$$ giving your answers to 2 decimal places.
AQA C4 2011 January Q6
10 marks
Standard +0.3
6
Given that \(\tan 2 x + \tan x = 0\), show that \(\tan x = 0\) or \(\tan ^ { 2 } x = 3\).
Hence find all solutions of \(\tan 2 x + \tan x = 0\) in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).
(l mark)
Given that \(\cos x \neq 0\), show that the equation $$\sin 2 x = \cos x \cos 2 x$$ can be written in the form $$2 \sin ^ { 2 } x + 2 \sin x - 1 = 0$$
Show that all solutions of the equation \(2 \sin ^ { 2 } x + 2 \sin x - 1 = 0\) are given by \(\sin x = \frac { \sqrt { 3 } - 1 } { p }\), where \(p\) is an integer.
AQA C4 2013 January Q3
12 marks
Standard +0.3
3
Express \(3 \cos x + 2 \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
(3 marks)
Hence find the minimum value of \(3 \cos x + 2 \sin x\) and the value of \(x\) in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\) where the minimum occurs. Give your value of \(x\) to the nearest \(0.1 ^ { \circ }\).
Show that \(\cot x - \sin 2 x = \cot x \cos 2 x\) for \(0 ^ { \circ } < x < 180 ^ { \circ }\).
Hence, or otherwise, solve the equation $$\cot x - \sin 2 x = 0$$ in the interval \(0 ^ { \circ } < x < 180 ^ { \circ }\).
AQA C4 2012 June Q2
7 marks
Standard +0.3
2
Express \(\sin x - 3 \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
Hence find the values of \(x\) in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\) for which $$\sin x - 3 \cos x + 2 = 0$$ giving your values of \(x\) to the nearest degree.
AQA C4 2013 June Q5
11 marks
Standard +0.3
5 The polynomial \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 4 x ^ { 3 } - 11 x - 3\).
Use the Factor Theorem to show that ( \(2 x + 3\) ) is a factor of \(\mathrm { f } ( x )\).
Write \(\mathrm { f } ( x )\) in the form \(( 2 x + 3 ) \left( a x ^ { 2 } + b x + c \right)\), where \(a , b\) and \(c\) are integers.
Show that the equation \(2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0\) can be written as \(4 x ^ { 3 } - 11 x - 3 = 0\), where \(x = \sin \theta\).
Hence find all solutions of the equation \(2 \cos 2 \theta \sin \theta + 9 \sin \theta + 3 = 0\) in the interval \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your solutions to the nearest degree.
AQA C4 2014 June Q5
15 marks
Standard +0.3
5
Express \(3 \sin x + 4 \cos x\) in the form \(R \sin ( x + \alpha )\) where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
Hence solve the equation \(3 \sin 2 \theta + 4 \cos 2 \theta = 5\) in the interval \(0 ^ { \circ } < \theta < 360 ^ { \circ }\), giving your solutions to the nearest \(0.1 ^ { \circ }\).
Show that the equation \(\tan 2 \theta \tan \theta = 2\) can be written as \(2 \tan ^ { 2 } \theta = 1\).
Hence solve the equation \(\tan 2 \theta \tan \theta = 2\) in the interval \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\), giving your solutions to the nearest \(0.1 ^ { \circ }\).
Use the Factor Theorem to show that \(2 x - 1\) is a factor of \(8 x ^ { 3 } - 4 x + 1\).
Show that \(4 \cos 2 \theta \cos \theta + 1\) can be written as \(8 x ^ { 3 } - 4 x + 1\) where \(x = \cos \theta\).
Given that \(\theta = 72 ^ { \circ }\) is a solution of \(4 \cos 2 \theta \cos \theta + 1 = 0\), use the results from parts (c)(i) and (c)(ii) to show that the exact value of \(\cos 72 ^ { \circ }\) is \(\frac { ( \sqrt { 5 } - 1 ) } { p }\) where \(p\) is an integer.
[0pt] [3 marks]
AQA C4 2015 June Q2
8 marks
Standard +0.3
2
Express \(2 \cos x - 5 \sin x\) in the form \(R \cos ( x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\), giving your value of \(\alpha\), in radians, to three significant figures.
Hence find the value of \(x\) in the interval \(0 < x < 2 \pi\) for which \(2 \cos x - 5 \sin x\) has its maximum value. Give your value of \(x\) to three significant figures.
Use your answer to part (a) to solve the equation \(2 \cos x - 5 \sin x + 1 = 0\) in the interval \(0 < x < 2 \pi\), giving your solutions to three significant figures.
[0pt] [3 marks]
Edexcel C4 Q8
13 marks
Standard +0.3
8. (i) Given that \(\cos ( x + 30 ) ^ { \circ } = 3 \cos ( x - 30 ) ^ { \circ }\), prove that tan \(x ^ { \circ } = - \frac { \sqrt { 3 } } { 2 }\).
(ii) (a) Prove that \(\frac { 1 - \cos 2 \theta } { \sin 2 \theta } \equiv \tan \theta\).
(b) Verify that \(\theta = 180 ^ { \circ }\) is a solution of the equation \(\sin 2 \theta = 2 - 2 \cos 2 \theta\).
(c) Using the result in part (a), or otherwise, find the other two solutions, \(0 < \theta < 360 ^ { \circ }\), of the equation using \(\sin 2 \theta = 2 - 2 \cos 2 \theta\).
Edexcel M2 Q6
15 marks
Standard +0.3
6. A particle \(P\) is projected from a point \(A\) on horizontal ground with speed \(u\) at an angle of elevation \(\alpha\) and moves freely under gravity. \(P\) hits the ground at the point \(B\).
Show that \(A B = \frac { u ^ { 2 } } { g } \sin 2 \alpha\). An archer fires an arrow with an initial speed of \(45 \mathrm {~ms} ^ { - 1 }\) at a target which is level with the point of projection and at a distance of 80 m . Given that the arrow hits the target,
find in degrees, correct to 1 decimal place, the two possible angles of projection.
Write down, with a reason, which of the two possible angles of projection would give the shortest time of flight.
(2 marks)
Show that the minimum time of flight is 1.8 seconds, correct to 1 decimal place.
(2 marks)
AQA FP1 2006 January Q3
5 marks
Moderate -0.8
3 Find the general solution, in degrees, for the equation $$\sin \left( 4 x + 10 ^ { \circ } \right) = \sin 50 ^ { \circ }$$
AQA FP1 2007 January Q7
8 marks
Moderate -0.3
7 The function f is defined for all real numbers by $$f ( x ) = \sin \left( x + \frac { \pi } { 6 } \right)$$
Find the general solution of the equation \(\mathrm { f } ( x ) = 0\).
The quadratic function g is defined for all real numbers by $$\mathrm { g } ( x ) = \frac { 1 } { 2 } + \frac { \sqrt { 3 } } { 2 } x - \frac { 1 } { 4 } x ^ { 2 }$$ It can be shown that \(\mathrm { g } ( x )\) gives a good approximation to \(\mathrm { f } ( x )\) for small values of \(x\).
Show that \(\mathrm { g } ( 0.05 )\) and \(\mathrm { f } ( 0.05 )\) are identical when rounded to four decimal places.
A chord joins the points on the curve \(y = \mathrm { g } ( x )\) for which \(x = 0\) and \(x = h\). Find an expression in terms of \(h\) for the gradient of this chord.
Using your answer to part (b)(ii), find the value of \(\mathrm { g } ^ { \prime } ( 0 )\).
AQA FP1 2009 January Q3
5 marks
Moderate -0.5
3 Find the general solution of the equation $$\tan \left( \frac { \pi } { 2 } - 3 x \right) = \sqrt { 3 }$$
AQA FP1 2011 January Q4
6 marks
Standard +0.3
4 Find the general solution of the equation $$\sin \left( 4 x - \frac { 2 \pi } { 3 } \right) = - \frac { 1 } { 2 }$$ giving your answer in terms of \(\pi\).
(6 marks)
AQA FP1 2012 January Q6
7 marks
Standard +0.3
6 Find the general solution of each of the following equations:
\(\quad \tan \left( \frac { x } { 2 } - \frac { \pi } { 4 } \right) = \frac { 1 } { \sqrt { 3 } }\);
\(\quad \tan ^ { 2 } \left( \frac { x } { 2 } - \frac { \pi } { 4 } \right) = \frac { 1 } { 3 }\).
AQA FP1 2013 January Q3
8 marks
Standard +0.3
3
Find the general solution of the equation $$\sin \left( 2 x + \frac { \pi } { 4 } \right) = \frac { \sqrt { 3 } } { 2 }$$ giving your answer in terms of \(\pi\).
Use your general solution to find the exact value of the greatest solution of this equation which is less than \(6 \pi\).
AQA FP1 2007 June Q6
6 marks
Moderate -0.3
6 Find the general solution of the equation $$\sin \left( 2 x - \frac { \pi } { 2 } \right) = \frac { \sqrt { 3 } } { 2 }$$ giving your answer in terms of \(\pi\).
AQA FP1 2008 June Q5
7 marks
Moderate -0.3
5
Find, in radians, the general solution of the equation $$\cos \left( \frac { x } { 2 } + \frac { \pi } { 3 } \right) = \frac { 1 } { \sqrt { 2 } }$$ giving your answer in terms of \(\pi\).
Hence find the smallest positive value of \(x\) which satisfies this equation.
AQA FP1 2009 June Q5
9 marks
Standard +0.3
5
Find the general solution of the equation $$\cos ( 3 x - \pi ) = \frac { 1 } { 2 }$$ giving your answer in terms of \(\pi\).
From your general solution, find all the solutions of the equation which lie between \(10 \pi\) and \(11 \pi\).
AQA FP1 2010 June Q3
5 marks
Moderate -0.3
3 Find the general solution, in degrees, of the equation $$\cos \left( 5 x - 20 ^ { \circ } \right) = \cos 40 ^ { \circ }$$
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AQA FP1 2011 June Q5
7 marks
Moderate -0.3
5
Find the general solution of the equation $$\cos \left( 3 x - \frac { \pi } { 6 } \right) = \frac { \sqrt { 3 } } { 2 }$$ giving your answer in terms of \(\pi\).
Use your general solution to find the smallest solution of this equation which is greater than \(5 \pi\).
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