MathematicsEasy45×since 2004Q5669Suppose θ∈[0,π4]\theta \in\left[0, \frac{\pi}{4}\right]θ∈[0,4π] is a solution of 4cosθ−3sinθ=14 \cos \theta-3 \sin \theta=14cosθ−3sinθ=1. Then cosθ\cos \thetacosθ is equal to :A6−6(36−2)\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}(36−2)6−6B4(36+2)\frac{4}{(3 \sqrt{6}+2)}(36+2)4C6+6(36+2)\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}(36+2)6+6D4(36−2)\frac{4}{(3 \sqrt{6}-2)}(36−2)4Check answerSkip