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6. (a) Solve, for -180° ≤ θ ≤ 180°, the equation 5 sin 2θ = 9 tan θ giving your answers, where necessary, to one decimal place - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 2
Question 8
6. (a) Solve, for -180° ≤ θ ≤ 180°, the equation 5 sin 2θ = 9 tan θ giving your answers, where necessary, to one decimal place. [Solutions based entirely on graph... show full transcript
Worked Solution & Example Answer:6. (a) Solve, for -180° ≤ θ ≤ 180°, the equation 5 sin 2θ = 9 tan θ giving your answers, where necessary, to one decimal place - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 2
Step 1
Solve the equation 5 sin 2θ = 9 tan θ
Answer
To solve the equation, we first need to rewrite it in a single trigonometric function format:
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Convert the equation using the identities for sine and tangent:
rac{5 sin(2θ)}{cos(θ)} = 9
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Simplify it to:
5 rac{2 sin(θ)cos(θ)}{cos(θ)} = 9
This leads to:
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Hence, we find:
sin(θ) = rac{9}{10}
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To find θ, we use the inverse sine:
θ = arcsinigg( rac{9}{10}igg)
Calculating gives us approximately:
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Since sin(θ) is positive in Quadrants I and II, we also consider:
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Thus, the solutions in the required interval are:
and (to one decimal place).
Step 2
Deduce the smallest positive solution to the equation 5 sin(2x - 50°) = 9 tan(x - 25°)
Answer
To solve the second part:
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Rewrite the equation:
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Use the definition of tangent:
tan(x - 25°) = rac{sin(x - 25°)}{cos(x - 25°)}
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Substitute back into the equation:
5 sin(2x - 50°) = 9 rac{sin(x - 25°)}{cos(x - 25°)}
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This requires solving for x, and utilizing identities, find:
(not within the positive range), but also:
as a valid smaller positive solution.