In a triangle $ABC, 3\cos A+2=0$. Then the quadratic equation with roots are $\sin A, \cos A$ is
- $9x^2 - 3x\left( \sqrt { 5 } -2 \right) -2\sqrt { 5 } =0$
- $9x^2 - 3x\left( \sqrt { 5 } - 2 \right) +2\sqrt { 5 } =0$
- $9x^2 - 3x\left( \sqrt { 7 } - 2 \right) +2\sqrt { 7 } =0$
- $9x^2 - 3x\left( \sqrt { 11 } - 2 \right) +2\sqrt { 11 }=0 $
3 cos A = -2, cos A = -2/3. sin A = sqrt(1 - 4/9) = sqrt(5)/3. Sum of roots = -2/3 + sqrt(5)/3 = (sqrt(5)-2)/3. Product = -2*sqrt(5)/9. Equation: x^2 - (sum)x + product = 0. x^2 - ((sqrt(5)-2)/3)x - 2*sqrt(5)/9 = 0. Multiply by 9: 9x^2 - 3x(sqrt(5)-2) - 2*sqrt(5) = 0.
Given 3cos A + 2 = 0, we find cos A equals -2/3, and since angle A is an angle of a triangle, sin A equals the positive square root of (1 - cos^2 A), which is square root of 5 divided by 3. The required quadratic equation is found by forming (x - sin A)(x - cos A) = 0, giving x^2 - (sin A + cos A)x + (sin A times cos A) = 0. Multiplying by 9 to clear the denominators results in the equation 9x^2 - 3x(square root of 5 - 2) - 2 times square root of 5 = 0.