Solve 4cos(x+50^(@))-sin(x-40^(@))=0 in the range x in 0 90^(@)). Give your answer in degrees.​

1 thought on “Solve 4cos(x+50^(@))-sin(x-40^(@))=0 in the range x in 0 90^(@)). Give your answer in degrees.​”

1. we have to solve 4cos(x + 50°) – sin(x – 40°) = 0 in the range of x ∈ [0° , 90°].

   solution : 4cos(x + 50°) – sin(x – 40°) = 0

   ⇒4co[(x – 40°) + 90°] – sin(x – 40°) = 0

   let x – 40° = A

   ⇒4cos(A + 90°) – sinA = 0

   we know, cos(90° + θ) = -sinθ

   ⇒-4sinA – sinA = 0

   ⇒-5sinA = 0

   ⇒sinA = 0

   ⇒A = nπ , where n belongs to all integers

   ⇒(x – 40°) = nπ

   for 0° < x < 90°

   we should take n = 0

   ⇒x – 40° = 0

   ⇒x = 40°

   Therefore the value of x is 40°.

   also read similar questions : please solve this trigonometry questionQ.

   Prove that : Sin x + Sin 3x+ Sin 5x +Sin 7x= 4cos x cos 2x sin4x

   Please so…

   https://brainly.in/question/4169823

   If sin a=1/2,show that (3vos a-4cos cube a) =0

   https://brainly.in/question/18824144

   Reply