Suppose ϵ (0 , 4 ) is a solution of 4 cos ⁡ -3 sin ⁡ =1. Then cos ⁡ is equal to :

Question

Accepted Answer

Correct answer: C. (c) 4 ( 1- tan^2 ⁡ /21+ tan^2 ⁡ /2 )-3 ( 2 tan ⁡ 21+ tan^2 ⁡ 2 )=1 let tan ⁡ 2=t 4-4t^2-6t1+t^2=14-4t^2-6t=1+t^2 ⇒5t^2+6t-3=0 ⇒t= -6 36-4 (5 ) (-3 )2 (5 ) = -6 9610 = -6 4 610t= -3+2 65 cos ⁡ = 1-t^21+t^2= 1- ( 2 6-35 )^21+ ( 2 6-35 )^2= 1- ( 24+9-12 625 )1+ ( 24+9-12 625 ) = 25-33+12 625+33-12 6= 12 6-858-12 6= 6 6-429-6 6 29+6 629+6 6 = 100+150 6625= 4+6 625 4-6 64-6 6 = -20025 (4-6 6 )= -84-6 6= 43 6-2

$Suppose θ ϵ (0, \frac{π}{4}) is a solution of 4 cos θ - 3 sin θ = 1 . Then cos θ is equal to :$

Solution

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