# Answer in Trigonometry for Oliver #245413

Find the **largest** value of parameter a such that equation 12sin^{2}x * cos^{2}x + (10a – 11)(cos^{4}x – sin^{4}x) = (2a-1)^{2} has at least one real solution.

**Solution:**

“12sin^2x times cos^2x + (10a – 11)(cos^4x – sin^4x) = (2a-1)^2n\ 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(2cos^2x -1)) = (2a-1)^2”

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

Put “cos x=y”

“\ 12(1-y^2) y^2 + (10a – 11)(2y^2 -1) = (2a-1)^2n\ Rightarrow12(1-y^2) y^2 + (10a – 11)(2y^2 -1) = (2a-1)^2n\ 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 = 0n\ Rightarrow6y^4 -10ay^2+5y^2+3a-5+2a^2 = 0n\ Rightarrow6y^4 +y^2(-10a+5)+3a-5+2a^2 = 0”

Put “y^2=t”

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

For at least one real solution, discriminant “Dge 0”

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

Now, “D ge 0”

“52a^2-172a+145ge0n\Rightarrow52left(a-frac{43}{26}right)^2+frac{36}{13}ge :0n\ Rightarrow 52left(a-frac{43}{26}right)^2ge :-frac{36}{13}n\ Rightarrow left(a-frac{43}{26}right)^2ge :-frac{9}{169}”

“mathrm{True:for:all}:a”

No largest value of ‘a’ is defined.