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Date May 2013 Marks available 3 Reference code 13M.2.hl.TZ2.6
Level HL only Paper 2 Time zone TZ2
Command term Solve Question number 6 Adapted from N/A

Question

Solve the equation \(3{\cos ^2}x - 8\cos x + 4 = 0\), where \(0 \leqslant x \leqslant 180^\circ \), expressing your answer(s) to the nearest degree.

[3]
a.

Find the exact values of \(\sec x\) satisfying the equation \(3{\sec ^4}x - 8{\sec ^2}x + 4 = 0\).

[3]
b.

Markscheme

attempting to solve for \(\cos x\) or for u where \(u = \cos x\) or for x graphically.     (M1)

EITHER

\(\cos x = \frac{2}{3}{\text{ (and 2)}}\)     (A1)

OR

\(x = 48.1897 \ldots ^\circ \)     (A1)

THEN

\(x = 48^\circ \)     A1

Note: Award (M1)(A1)A0 for \(x = 48^\circ ,{\text{ }}132^\circ \).

 

Note: Award (M1)(A1)A0 for 0.841 radians.

 

[3 marks]

a.

attempting to solve for \(\sec x\) or for \(v\) where \(v = \sec x\).     (M1)

\(\sec x = \pm \sqrt 2 {\text{ }}\left( {{\text{and }} \pm \sqrt {\frac{2}{3}} } \right)\)     (A1)

\(\sec x = \pm \sqrt 2 \)     A1

[3 marks]

b.

Examiners report

Part (a) was generally well done. Some candidates did not follow instructions and express their final answer correct to the nearest degree. A large number of candidates successfully employed a graphical approach.

a.

Part (b) was not well done. Common errors included attempting to solve for x rather than for \(\sec x\), either omitting or not considering \(\sec x =  - \sqrt 2 \), not rejecting \(\sec x =  \pm \sqrt {\frac{2}{3}} \) and not working with exact values.

b.

Syllabus sections

Topic 3 - Core: Circular functions and trigonometry » 3.6 » Algebraic and graphical methods of solving trigonometric equations in a finite interval, including the use of trigonometric identities and factorization.
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