lo g sin ( x ) cos ( x ) + lo g cos ( x ) sin ( x ) = 2
If the minimum value of x that satisfy the equation above is equals to a π for some constant a , find the value of a .
Give your answer to 3 decimal places.
Note: Angles are measured in radians.
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Well done using the properties of logarithm. Bonus question: What would be the solution of all x such that the equation is fulfilled?
@Calvin Lin /@mathchallengemaster it would be n π + 4 π
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This will be incorrect , remember that the for lo g a b b ∈ ( 0 , ∞ ) − { 1 } and a ∈ ( 0 , ∞ )
Both sin x and cos x should be positive , which means x should lie in the 1st Quadrant
Haha We posted identical solutions at the exact same time. No point in leaving mine up any more. :)
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Let lo g sin ( x ) cos ( x ) = m
∴ lo g cos ( x ) sin ( x ) = m 1
⟹ m + m 1 = 2
m 2 + 1 = 2 m ⟹ ( m − 1 ) 2 = 0 , ∴ m = 1
So, sin ( x ) = cos ( x ) ⟹ x ∣ m i n = 4 π
So a = 4 .