If $$6{\sin ^4}\theta + 3{\cos ^4}\theta = 2{\text{,}}$$     then the value of $${\left[ {7{{\operatorname{cosec} }^6}\theta + 8{{\sec }^6}\theta } \right]^{\frac{1}{3}}}$$     is?

Solution(By Examveda Team)

6sin⁴θ + 3cos⁴θ = 2
⇒ 6sin⁴θ + 3(1 - sin²θ )² = 2
⇒ 6sin⁴θ + 3 + 3sin⁴θ - 6sin²θ = 2
⇒ 9sin⁴θ - 6sin²θ + 1 = 0
⇒ (3sin²θ -1)² = 0
⇒ 3sin²θ = 1
⇒ sinθ = $$\frac{1}{{\sqrt 3 }} $$


Now, $${\left[ {7{{\operatorname{cosec} }^6}\theta + 8{{\sec }^6}\theta } \right]^{\frac{1}{3}}}$$
$$\eqalign{ & = {\left[ {7 \times {{(\sqrt 3 )}^6} + 8{{\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)}^6}} \right]^{\frac{1}{3}}} \cr & = {\left[ {7 \times 27 + 8 \times \frac{{27}}{8}} \right]^{\frac{1}{3}}} \cr & = {\left( {8 \times 27} \right)^{\frac{1}{3}}} \cr & = 6 \cr} $$