Scottish Highers Maths Addition Formula: 15. (a) Solve the equation sin 2

15. (a) Solve the equation sin 2x² + 6cos x² = 0 for 0° < x < 360° - Scottish Highers Maths - Question 15 - 2019

Question 15

15. (a) Solve the equation sin 2x² + 6cos x² = 0 for 0° < x < 360°. (b) Hence solve sin 4x² + 6cos 2x² = 0 for 0° < x < 360°.

Worked Solution & Example Answer:15. (a) Solve the equation sin 2x² + 6cos x² = 0 for 0° < x < 360° - Scottish Highers Maths - Question 15 - 2019

Step 1

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Answer

To solve the equation, we start by substituting the double angle formula for sine into the equation:

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

Substituting this gives:

$2 ext{sin}(x) ext{cos}(x) + 6 ext{cos}(x) = 0$

Factoring out the common term:

$ext{cos}(x)(2 ext{sin}(x) + 6) = 0$

This gives us two equations to solve:

Thus, the only valid solutions from part (a) are:

Step 2

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Answer

Using the double angle formula for cosine, we have:

$ext{cos}(2x) = 2 ext{cos}^2(x) - 1$

And substituting we get:

$ext{sin}(4x) + 6(2 ext{cos}^2(x) - 1) = 0$

From part (a), we know:

$ext{sin}(4x) = 0\ ext{solutions: } 0°, 180° ext{ (and taking into account the initial conditions)}$

Thus the valid solutions for part (b) are: