( √3 - 1 )sin θ + ( √3 + 1 )cos θ = 2
Dividing by 2√2 on both sides
( √3 - 1 ) /2√2 . sin θ + ( √3 + 1 ) /2√2 . cos θ = 2/2√2
We know that
Substitute the trigonometric ratios in the place of values
sin θ. sin π/12 + cos θ. cos π/12 = 1/√2
Using cos ( a - b) = cos a. cos b + sin a. sin b
cos ( θ - π/12 ) = 1/√2
cos ( θ - π/12 ) = cos π/4
If cos θ = cos a, a is the principal solution,
General solution is θ = 2nπ ± a where a Belongs to ( 0, π ]
Therefore,
θ - π/12 = 2nπ ± π/4
θ = 2nπ ± π/4 + π/12
Hence we get the general solution as θ = 2nπ ± π/4 + π/12 ( A) Option.
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Trigonometric equations
( √3 - 1 )sin θ + ( √3 + 1 )cos θ = 2
Dividing by 2√2 on both sides
( √3 - 1 ) /2√2 . sin θ + ( √3 + 1 ) /2√2 . cos θ = 2/2√2
We know that
Substitute the trigonometric ratios in the place of values
sin θ. sin π/12 + cos θ. cos π/12 = 1/√2
Using cos ( a - b) = cos a. cos b + sin a. sin b
cos ( θ - π/12 ) = 1/√2
cos ( θ - π/12 ) = cos π/4
We know that
If cos θ = cos a, a is the principal solution,
General solution is θ = 2nπ ± a where a Belongs to ( 0, π ]
Therefore,
θ - π/12 = 2nπ ± π/4
θ = 2nπ ± π/4 + π/12
Hence we get the general solution as θ = 2nπ ± π/4 + π/12 ( A) Option.