The value of K for which the 2cos4x - sin4x +k=0 has at least one solution can be?

Question

The value of K for which  the 2cos⁴x - sin⁴x +k=0 has at least one solution can be?

Answer 1

2?cos?^4 -?sin?^4 x+k=0
2?cos?^4 -(1-?cos?^2 x)(1-?cos?^2 x)+k=0
2?cos?^4 -(1-2?cos?^2 x+?cos?^4 x)+k=0
?cos?^4 +2?cos?^2 x+k-1=0
Substituting y=?cos?^2 
y^2+2y+k-1=0
Solving the quadratic equation:
(-2±√(4-4(k+1)))/2
?cos?^2 = -1±√(2-k)
Because of the square root k must be k <=2, but ?cos?^2  ≥0 therefore k ≤1 and only the positive square root produces valid outcome.
If k = 1 ->?cos?^2 x=0 therefore x = pi/2,3pi/2,etc

 

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George