Exams Joint Entrance Examination Mathematics Trigonometric Ratios and Identities Question 17 Question 17 of 298 Medium MCQ
Suppose θ ∈ [ 0 , π 4 ] \theta \in [0,\frac{\pi }{4}] θ ∈ [ 0 , 4 π ] is a solution of 4 cos θ − 3 sin θ = 1 4\cos \theta -3\sin \theta =1 4 cos θ − 3 sin θ = 1 . Then cos θ \cos \theta cos θ is equal to :
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\[ 4 cos θ − 3 sin θ = 1 4 cos θ − 1 = 3 sin θ 16 cos 2 θ + 1 − 8 cos θ = 9 ( 1 − cos 2 θ ) ⇒ 25 cos 2 θ − 8 cos θ − 8 = 0 ⇒ cos θ = 8 ± 64 + 4 × 25 × 8 2.25 = 8 ± 4 4 + 50 2.25 \begin{aligned}
& 4 \cos \theta-3 \sin \theta=1 \\
& 4 \cos \theta-1=3 \sin \theta \\
& 16 \cos ^2 \theta+1-8 \cos \theta=9\left(1-\cos ^2 \theta\right) \\
& \Rightarrow 25 \cos ^2 \theta-8 \cos \theta-8=0 \\
& \Rightarrow \cos \theta=\frac{8 \pm \sqrt{64+4 \times 25 \times 8}}{2.25} \\
& =\frac{8 \pm 4 \sqrt{4+50}}{2.25}
\end{aligned} 4 cos θ − 3 sin θ = 1 4 cos θ − 1 = 3 sin θ 16 cos 2 θ + 1 − 8 cos θ = 9 ( 1 − cos 2 θ ) ⇒ 25 cos 2 θ − 8 cos θ − 8 = 0 ⇒ cos θ = 2.25 8 ± 64 + 4 × 25 × 8 = 2.25 8 ± 4 4 + 50 \]
\[ = 4 ± 2 54 25 As θ ∈ [ 0 , π 4 ] ⇒ cos θ = 4 + 6 6 25 = 4 3 6 − 2 \begin{aligned}
& =\frac{4 \pm 2 \sqrt{54}}{25} \\
& \text { As } \theta \in\left[0, \frac{\pi}{4}\right] \\
& \Rightarrow \cos \theta=\frac{4+6 \sqrt{6}}{25}=\frac{4}{3 \sqrt{6}-2}
\end{aligned} = 25 4 ± 2 54 As θ ∈ [ 0 , 4 π ] ⇒ cos θ = 25 4 + 6 6 = 3 6 − 2 4 \] Answer the question to unlock