Solve
cosec² 2x - cot 2x = 1
for 0 ≤ x ≤ 180°. - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2
Question 9
Solve
cosec² 2x - cot 2x = 1
for 0 ≤ x ≤ 180°.
Worked Solution & Example Answer:Solve
cosec² 2x - cot 2x = 1
for 0 ≤ x ≤ 180°. - Edexcel - A-Level Maths Pure - Question 9 - 2010 - Paper 2
Using cosec² 2x - cot 2x = 1
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We start with the equation:
extcosec22x−extcot2x=1.
Using the identity for cosecant, we can rewrite the equation as:
1+extcot22x−extcot2x=1.
Simplifying this gives us:
extcot22x−extcot2x=0.Factoring the Equation
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We can factor this expression:
extcot2x(extcot2x−1)=0.
Setting each factor to zero gives us two equations to solve:
extcot2x=0,
extcot2x−1=0.Solving cot 2x = 0
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For the first equation,
extcot2x=0,
we find that this occurs when:
2x=90°+kimes180°,extwherekextisaninteger.
For the range 0 ≤ x ≤ 180°, we find:
2x=90°ightarrowx=45°,
2x=270°ightarrowx=135°.Solving cot 2x - 1 = 0
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For the second equation:
extcot2x=1,
this occurs when:
2x=45°+kimes180°,extwherekextisaninteger.
In the range 0 ≤ x ≤ 180°, we derive:
2x=45°ightarrowx=22.5°,
2x=225°ightarrowx=112.5°.Final Solutions
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Combining all solutions obtained, we have:
x=22.5°,45°,112.5°,135°.
Thus, the final answers are:
extBothx=22.5°extandx=112.5°.Join the A-Level students using SimpleStudy...
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