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 graphical or numerical methods are not acceptable.] (b) Deduce the smallest positive solution to the equation 5 sin(2x - 50°) = 9 tan(x - 25°)

A-Level Edexcel Maths Pure Simultaneous Equations: 6. (a) Solve, for -180° ≤ θ ≤ 18

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 7 - 2019 - Paper 2

Question 7

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 7 - 2019 - Paper 2

Step 1

Solve, for -180° ≤ θ ≤ 180°, the equation 5 sin 2θ = 9 tan θ

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Answer

To solve the equation, we start by rewriting it in a simpler form. We know:

$5 ext{sin}(2 heta) = 9 ext{tan}( heta)$

Using the identity for sin(2θ), we have:

$ext{sin}(2 heta) = 2 ext{sin}( heta) ext{cos}( heta)$

Substituting this into the equation gives:

$5(2 ext{sin}( heta) ext{cos}( heta)) = 9 rac{ ext{sin}( heta)}{ ext{cos}( heta)}$

Simplifying, we get:

$10 ext{sin}( heta) ext{cos}( heta) = rac{9 ext{sin}( heta)}{ ext{cos}( heta)}$

Assuming ( ext{sin}( heta) \neq 0 ), we can divide both sides by ( ext{sin}( heta) ) leading to:

$10 ext{cos}^2( heta) = 9$

From this, we have:

$ext{cos}^2( heta) = 0.9$

Taking the square root:

$ext{cos}( heta) = rac{3}{ ext{sqrt}(10)} ext{ or } ext{cos}( heta) = - rac{3}{ ext{sqrt}(10)}$

Now, we find θ for both cases. Using the arccosine function:

For ( ext{cos}( heta) = rac{3}{ ext{sqrt}(10)} ):

One solution is ( heta = 18.4° )
The second solution, considering the cosine function is even, is ( heta = -18.4° )

For ( ext{cos}( heta) = -rac{3}{ ext{sqrt}(10)} ):

This provides solutions in the second and third quadrants:
- The solution in the second quadrant is ( heta = 180° - 18.4° = 161.6° )
- The solution in the third quadrant is ( heta = 180° + 18.4° = 198.4° ) (not in range)

Thus, valid solutions are ( heta = 18.4°, -18.4°, 161.6° ).

Step 2

Deduce the smallest positive solution to the equation 5 sin(2x - 50°) = 9 tan(x - 25°)

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Answer

To find the smallest positive solution, we analyze the provided equation:

$5 ext{sin}(2x - 50°) = 9 ext{tan}(x - 25°)$

Using the identity for sin(2x), we can rewrite it as:

$ext{sin}(2x - 50°) = 2 ext{sin}(x-25°) ext{cos}(x-25°)$

Thus, substituting, we get:

$10 ext{sin}(x-25°) ext{cos}(x-25°) = 9 rac{ ext{sin}(x-25°)}{ ext{cos}(x-25°)}$

Assuming ( ext{sin}(x-25°) \neq 0 ) (as above), we find:

Simplify:

$10 ext{cos}^2(x-25°) = 9$

Therefore,

$ext{cos}^2(x-25°) = 0.9$

Taking the square root:

$ext{cos}(x-25°) = rac{3}{ ext{sqrt}(10)} \text{ or } - rac{3}{ ext{sqrt}(10)}$

For positive ( x ):

When ( ext{cos}(x-25°) = rac{3}{ ext{sqrt}(10)} ):
- We calculate: ( x - 25° = 18.4° ) ( x = 43.4° )
When ( ext{cos}(x-25°) = -rac{3}{ ext{sqrt}(10)} ):
- Solution in second quadrant gives:
- ( x - 25° = 180° - 18.4° )
- So, ( x = 180° - 18.4° + 25° = 186.6° ) (not valid, needs to be smallest)

Thus, the smallest positive solution is:

$x = 43.4°$ .

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 7 - 2019 - Paper 2

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 7 - 2019 - Paper 2

Solve, for -180° ≤ θ ≤ 180°, the equation 5 sin 2θ = 9 tan θ

Deduce the smallest positive solution to the equation 5 sin(2x - 50°) = 9 tan(x - 25°)

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