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8. (i) Solve, for −180° ≤ x < 180°, tan(x − 40°) = 1.5 giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 10 - 2013 - Paper 4

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8. (i) Solve, for −180° ≤ x < 180°, tan(x − 40°) = 1.5 giving your answers to 1 decimal place. (ii) (a) Show that the equation sin θ tan θ = 3 cos θ + 2 can be ... show full transcript

Worked Solution & Example Answer:8. (i) Solve, for −180° ≤ x < 180°, tan(x − 40°) = 1.5 giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 10 - 2013 - Paper 4

Step 1

Solve, for −180° ≤ x < 180°, tan(x − 40°) = 1.5

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Answer

To solve the equation, we first find the general solution for the tangent function:

Recall that $an(x) = rac{ ext{opposite}}{ ext{adjacent}}$ .
Use the arctangent function to solve for the angle. $x - 40° = an^{-1}(1.5)$
Calculate the angle: $an^{-1}(1.5) ext{ gives approximately } 56.3099°.$
Now add the periodicity of the tangent function, which is 180°: $x - 40° = 56.3099° + k imes 180°$ where k is an integer.
Solve for x: $x = 56.3099° + 40° + k imes 180°$ $x = 96.3099° + k imes 180°$ For k = 0, $x ext{ is } 96.3° ext{ (1 decimal place)}$ For k = 1, $x = 96.3099° + 180° = 276.3099°$ (not in the interval). Thus, the only solution is: x ≈ 96.3°.

Step 2

Show that the equation sin θ tan θ = 3 cos θ + 2 can be written in the form 4 cos² θ + 2 cos θ − 1 = 0

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Answer

To transform the equation, start with the given expression:

Rewrite tan θ as sin θ/cos θ:
$rac{ ext{sin }θ}{ ext{cos }θ}$
Thus, $ext{sin }θ rac{ ext{sin }θ}{ ext{cos }θ} = 3 ext{cos }θ + 2$
This simplifies to $rac{ ext{sin }^2 θ}{ ext{cos }θ} = 3 ext{cos }θ + 2$
Multiply through by cos θ to eliminate the fraction: $ext{sin }^2 θ = 3 ext{cos }^2 θ + 2 ext{cos }θ$
Substitute $ext{sin }^2 θ = 1 - ext{cos }^2 θ$ to get: $1 - ext{cos }^2 θ = 3 ext{cos }^2 θ + 2 ext{cos }θ$
Rearranging gives: $1 - 4 ext{cos }^2 θ - 2 ext{cos }θ = 0$ Thus, we have: 4 cos² θ + 2 cos θ − 1 = 0.

Step 3

Hence solve, for 0 ≤ θ < 360°, sin θ tan θ = 3 cos θ + 2

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Answer

To solve the equation derived earlier:

Starting from 4 cos² θ + 2 cos θ − 1 = 0: This is a quadratic in cos θ. Let $u = ext{cos }θ$ : $4u^2 + 2u - 1 = 0$
Applying the quadratic formula: $u = rac{-b ext{ ± } ext{√}(b^2 - 4ac)}{2a}$ where a = 4, b = 2, c = -1. Thus: $u = rac{-2 ext{ ± } ext{√}(2^2 - 4(4)(-1))}{2(4)}$ $u = rac{-2 ext{ \pm } ext{\sqrt}(4 + 16)}{8}$ $u = rac{-2 ext{ \pm } ext{\sqrt}20}{8}$ $u = rac{-2 ext{ \pm } 2 ext{\sqrt5}}{8}$ $u = rac{-1 ext{ \pm } ext{\sqrt5}}{4}$
The two potential solutions for cos θ are: $u_1 = rac{-1 + ext{√5}}{4}$ $u_2 = rac{-1 - ext{√5}}{4}$
Evaluate these:
- For u_1:
  - Calculate approximate value and find possible angles.
- For u_2 (being negative) will lead to a complex solution outside of our range.
Solutions will yield angles 72°, 144°, and possible repeat in 288°. Thus, the solutions are: θ = 72°, 144°, and 288°.

8. (i) Solve, for −180° ≤ x < 180°, tan(x − 40°) = 1.5 giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 10 - 2013 - Paper 4

Worked Solution & Example Answer:8. (i) Solve, for −180° ≤ x < 180°, tan(x − 40°) = 1.5 giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 10 - 2013 - Paper 4

Solve, for −180° ≤ x < 180°, tan(x − 40°) = 1.5

Show that the equation sin θ tan θ = 3 cos θ + 2 can be written in the form 4 cos² θ + 2 cos θ − 1 = 0

Hence solve, for 0 ≤ θ < 360°, sin θ tan θ = 3 cos θ + 2

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