12. (a) Solve, for -180° < x < 180°, the equation 3 sin²x + sin x + 8 = 9 cos²x (b) Hence find the smallest positive solution of the equation 3sin(20° - 30°) + sin(20° - 30°) + 8 = 9 cos(20° - 30°) - Edexcel - A-Level Maths Pure - Question 13 - 2017 - Paper 2

To solve the equation, we can use the identity for cosine:

$ext{cos}^2 x = 1 - ext{sin}^2 x$

Substituting this into the equation gives:

$3 ext{sin}^2 x + ext{sin} x + 8 = 9(1 - ext{sin}^2 x)$

Now we rearrange the equation:

$3 ext{sin}^2 x + ext{sin} x + 8 = 9 - 9 ext{sin}^2 x$

Combining like terms results in:

$12 ext{sin}^2 x + ext{sin} x - 1 = 0$

We now have a quadratic in terms of $ext{sin} x$, which we can solve using the quadratic formula:

ext{sin} x = rac{-b oxed{ ext{±}} ext{sqrt}(b^2 - 4ac)}{2a}

Where

• $a = 12$, $b = 1$, and $c = -1$.

Calculating the discriminant:

$b^2 - 4ac = 1^2 - 4(12)(-1) = 1 + 48 = 49$

Substituting back, we find:

ext{sin} x = rac{-1 ext{±} 7}{24}

This gives us two values:

1. ext{sin} x = rac{6}{24} = rac{1}{4}
2. ext{sin} x = rac{-8}{24} = -rac{1}{3}

Using the inverse sine function to find the angles:

• For ext{sin} x = rac{1}{4}:

  • $x_1 = 14.48°$
  • $x_2 = 165.52°$

• For ext{sin} x = -rac{1}{3}:

  • $x_3 = -19.47°$
  • $x_4 = -160.53°$

Thus, the solutions are:

• $x = 14.48°, 165.52°, -19.47°, -160.53°$