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Question

Mathematics Question on Trigonometric Functions

If sin⁡2 θ+3 cos⁡ θ−2=0\sin^2 \, \theta + 3 \, \cos \, \theta - 2 = 0, then cos⁡3 θ+sec⁡3 θ\cos^3 \, \theta + \sec^3 \, \theta is equal to

A

18

B

9

C

4

D

14\frac{1}{4}

Answer

18

Explanation

Solution

Given sin⁡2 θ+3 cos⁡ θ−2=0\sin^2 \, \theta + 3 \, \cos \, \theta - 2 = 0 ⇒cos⁡2θ−3cos⁡θ+1=0\Rightarrow \cos^{2} \theta - 3 \cos \theta + 1 = 0 ⇒cos⁡2θ+1=3cos⁡θ\Rightarrow \cos^{2} \theta +1 = 3 \cos\theta ⇒cos⁡θ+1cos⁡θ=3,\Rightarrow \cos\theta + \frac{1}{\cos\theta} = 3 , cubing both sides, we get cos⁡3θ+1cos⁡3θ+3(cos⁡θ+1cos⁡θ)=27 \cos^{3} \theta + \frac{1}{\cos^{3} \theta} + 3 \left(\cos \theta + \frac{1}{\cos\theta} \right) = 27 ⇒cos⁡3θ+sec⁡3θ=27−3×3=18\Rightarrow \cos^{3} \theta + \sec^{3} \theta = 27 - 3 \times3 = 18