How many solutions on the interval [0, 2020π] does the equation sin 2x + 1 = sin x + cos x have?
Answers
Given : sin 2x + 1 = sin x + cos x
To Find : number of solutions on the interval [0, 20π]
Solution:
sin 2x + 1 = sin x + cos x
=> 2sinxcosx + sin²x + cos²x = sin x + cos x
=> (sinx + cosx)² = (sinx + cosx)
=> (sinx + cosx) ( sinx + cosx - 1) = 0
=> sinx + cosx = 0 or sinx + cosx - 1 = 0
sinx + cosx = 0
Dividing both sides by cosx Hence
=> tanx = -1 => x = ( nπ - π/4)
sinx + cosx - 1 = 0
=> sinx + cosx = 1
=> √2 (cosx/√2 + sinx/√2 ) = 1
=> √2( cosxCosπ/4 + sinx sinπ/4 ) = 1
cos ( A- B) = CosA CosB + SinASinB
=> √2 cos (x - π/4) = 1
=> cos (x - π/4) = 1/√2
1/√2 = cosπ/4
=> x - π/4 = 2nπ ± π/4
=> x = 2nπ or 2nπ + π/2
x = 2nπ => 11 solutions from [0 , 20π] ( as 0 and 20π both included)
x = 2nπ + π/2 => 10 solutions from [0 , 20π] ( 1 solution in Every 2π )
x = ( nπ - π/4) => 20 solutions from [0 , 20π] ( 2 solutions in Every 2π )
Hence 41 solutions
Learn More:
1 (2x-1/2x+1)=tan^-1 (23/36)
brainly.in/question/8925658
tan x + sec x = √3 find the value of x - Brainly.in
brainly.in/question/4638354