<s>[INST] If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$ If $\sin^2 \theta + 2\cos \theta – 2 = 0$, then find the value of $\cos^3 \theta + \sec^3 \theta$.
 [/INST] Hint: Just use $\sin^2\theta = 1 - \cos^2\theta$, get the eqn in $(a-b)^2 = 0$ form. 
Then, simply use the fact that $\sec\theta = \frac1{\cos\theta}$
Spoiler :
!> $$\sin^2 \theta + 2\cos\theta - 2 = 0 \\
\implies 1 - \cos^2\theta + 2\cos\theta -2 = 0\\
\implies -(\cos^2\theta - 2 \cos\theta + 1^2) = 0\\
\implies (\cos\theta - 1)^2 = 0\\
\implies \cos\theta = 1\\
\implies \cos^3 \theta = 1^3 \quad\dots(1)\\
\implies \sec^3\theta = 1 \quad\dots(2)$$
From $(1)$ and $(2)$, the answer is $1+1$
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