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7. (i) Solve, for $0 eq x < rac{eta}{2}$, the equation 4sin$x = sec$x - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2
Question 9
7. (i) Solve, for $0 eq x < rac{eta}{2}$, the equation 4sin$x = sec$x. (ii) Solve, for $0 eq heta < 360^{ extcirc}$, the equation 5sin$ heta - 5cos heta = 2$ g... show full transcript
Worked Solution & Example Answer:7. (i) Solve, for $0 eq x < rac{eta}{2}$, the equation 4sin$x = sec$x - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2
Step 1
Solve, for $0 eq x < rac{eta}{2}$, the equation 4sin$x = sec$x.
Answer
To solve the equation, we start with:
Using the identity for secant, we can rewrite this as:
4 ext{sin}x = rac{1}{ ext{cos}x}
Multiplying both sides by (ensuring ), we get:
This can be transformed with the double angle identity:
Thus, we have:
ext{sin}(2x) = rac{1}{2}
The solutions for ext{sin}(2x) = rac{1}{2} are:
2x = rac{eta}{6} + 2keta ext{ and } 2x = rac{5eta}{6} + 2keta
Solving for , we get:
x = rac{eta}{12} + keta ext{ and } x = rac{5eta}{12} + keta
Considering the interval 0 eq x < rac{eta}{2}, we find:
- For :
- x = rac{eta}{12} ext{ and } x = rac{5eta}{12}
- Both values are within the range.
Step 2
Solve, for $0 eq heta < 360^{ extcirc}$, the equation 5sin$ heta - 5cos heta = 2$
Answer
Starting with the equation:
We can rearrange this to isolate terms involving sin and cos:
Dividing both sides by 5 gives:
ext{sin} heta - ext{cos} heta = rac{2}{5}
Expressing and in terms of ext{Rsin}( heta - eta), we identify and eta with:
- aneta = rac{1}{1} = 1 thus eta = 45^{ extcirc}
The equation now reads:
ext{√2sin}( heta - 45^{ extcirc}) = rac{2}{5}
Solving for results in:
heta - 45^{ extcirc} = ext{arcsin} rac{2}{5 ext{√2}}
Calculating gives two angles in the specified range, leading to:
- heta = 45^{ extcirc} + ext{arcsin} rac{2}{5 ext{√2}} ext{ (1)}
- heta = 225^{ extcirc} + ext{arcsin} rac{2}{5 ext{√2}} ext{ (2)}
Finally, approximating both angles gives answers to one decimal place.