Explanation

Using sin⁶θ+cos⁶θ = 1−3sin²θcos²θ and sin⁴θ+cos⁴θ = 1−2sin²θcos²θ: 2(1−3s²c²)−3(1−2s²c²) = −1. A quick check: at θ = 0 the expression is 2(1)−3(1) = −1, and it is constant in θ. The keyed option is A.