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8. (a) Show that the equation 4sin²x + 9cosx - 6 = 0 can be written as 4cos²x - 9cosx + 2 = 0 - Edexcel - A-Level Maths: Pure - Question 10 - 2009 - Paper 2
Question 10
8. (a) Show that the equation 4sin²x + 9cosx - 6 = 0 can be written as 4cos²x - 9cosx + 2 = 0. (2) (b) Hence solve, for 0 ≤ x < 720°, 4sin²x + 9cosx - 6 = 0, g... show full transcript
Worked Solution & Example Answer:8. (a) Show that the equation 4sin²x + 9cosx - 6 = 0 can be written as 4cos²x - 9cosx + 2 = 0 - Edexcel - A-Level Maths: Pure - Question 10 - 2009 - Paper 2
Step 1
Show that the equation can be written as 4cos²x - 9cosx + 2 = 0.
Answer
To rewrite the equation from part (a), we start with the original equation:
Using the Pythagorean identity, we know that:
Substituting this identity into the original equation gives:
Expanding this, we have:
Simplifying this results in:
Multiplying the entire equation by -1 for clarity, we obtain:
Thus, we have shown that:
can be expressed as:
Step 2
Hence solve, for 0 ≤ x < 720°, 4sin²x + 9cosx - 6 = 0.
Answer
Using the equation from part (a):
This is a quadratic equation in terms of . Letting , we rewrite it as:
Applying the quadratic formula, , with , , and :
Calculating the discriminant:
Thus,
Calculating the two potential solutions:
- (not valid, since must be between -1 and 1)
So, .
Now, to find the angles:
Using which is approximately .
Since is positive in the first and fourth quadrants:
- First solution:
- Second solution:
- Third solution for the next cycle:
- Fourth solution:
Now, we also include to check:
- (not in range)
- (not in range)
Therefore, the valid solutions for to 1 decimal place are: