MHT CET · Maths · Trigonometric Equations
If \(2 \sin ^{2} x+7 \cos x=5\), then permissible value of \(\cos x\) is
- A \(\frac{1}{2}\)
- B 0
- C 1
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
We have \(2 \cos ^{2} \theta+3 \cos \theta=2\)
\(2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0 \Rightarrow 2 \cos \theta(\cos \theta+2)-\)\(1(\cos \theta+2)=0\)
\((2 \cos \theta-1)(\cos \theta+2)=0\)
\(\therefore \cos \theta=\frac{1}{2},-2\) (Impossible) \(\Rightarrow \cos \theta=\frac{1}{2}\)
\(2 \cos ^{2} \theta+4 \cos \theta-\cos \theta-2=0 \Rightarrow 2 \cos \theta(\cos \theta+2)-\)\(1(\cos \theta+2)=0\)
\((2 \cos \theta-1)(\cos \theta+2)=0\)
\(\therefore \cos \theta=\frac{1}{2},-2\) (Impossible) \(\Rightarrow \cos \theta=\frac{1}{2}\)
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