Answer to Question #245413 in Trigonometry for Oliver

Question

Find the largest value of parameter a such that equation 12sin²x * cos²x + (10a - 11)(cos⁴x - sin⁴x) = (2a-1)² has at least one real solution.

Answer 1

Solution:

$12\sin^2x \times \cos^2x + (10a - 11)(\cos^4x - \sin^4x) = (2a-1)^2 \\ \Rightarrow 12(1-\cos^2x) \times \cos^2x + (10a - 11)((\cos^2x)^2 - (\sin^2x)^2) = (2a-1)^2$

$\\ \Rightarrow 12(1-\cos^2x) \times \cos^2x + (10a - 11)((\cos^2x+\sin^2x)(\cos^2x - \sin^2x)) = (2a-1)^2$

$\\ \Rightarrow 12(1-\cos^2x) \times \cos^2x + (10a - 11)(1(2\cos^2x -1)) = (2a-1)^2$

$\\ \Rightarrow 12(1-\cos^2x) \cos^2x + (10a - 11)(2\cos^2x -1) = (2a-1)^2$

Put $\cos x=y$

$\\ 12(1-y^2) y^2 + (10a - 11)(2y^2 -1) = (2a-1)^2 \\ \Rightarrow12(1-y^2) y^2 + (10a - 11)(2y^2 -1) = (2a-1)^2 \\ \Rightarrow12y^2-12y^4 + 20ay^2-22y^2-10a+11 = 4a^2+1-4a$

$\\ \Rightarrow-12y^4 + 20ay^2-10y^2-6a+10-4a^2 = 0 \\ \Rightarrow6y^4 -10ay^2+5y^2+3a-5+2a^2 = 0 \\ \Rightarrow6y^4 +y^2(-10a+5)+3a-5+2a^2 = 0$

Put $y^2=t$

$6t^2 +t(-10a+5)+3a-5+2a^2 = 0 \\\Rightarrow6y^4 +y^2(-10a+5)+3a-5+2a^2 = 0$

For at least one real solution, discriminant $D\ge 0$

$D=B^2-4AC=(-10a+5)^2-4(6)(3a-5+2a^2) \\=100a^2+25-100a-24(3a-5+2a^2) \\=100a^2+25-100a-72a+120-48a^2 \\=52a^2-172a+145$

Now, $D \ge 0$

$52a^2-172a+145\ge0 \\\Rightarrow52\left(a-\frac{43}{26}\right)^2+\frac{36}{13}\ge \:0 \\ \Rightarrow 52\left(a-\frac{43}{26}\right)^2\ge \:-\frac{36}{13} \\ \Rightarrow \left(a-\frac{43}{26}\right)^2\ge \:-\frac{9}{169}$

$\mathrm{True\:for\:all}\:a$

No largest value of 'a' is defined.