Question

12 cos tetha minus 16 sin theta is equal to 0 find 2 sin theta plus cos theta​

Answer 1

12 cos θ - 16 sinθ = 0

=> 3 cos θ -4 sin θ = 0

=> tanθ = 3/4

let P = 3x

B = 4x

=> H= 5x

=> sinθ = P/H = 3/5

cosθ = B/H = 4/5

2sinθ + cosθ

= 2× 3/5 + 4/5

= 6/5+4/5

= 10/5

= 2

=> 2sinθ + cosθ = 2

Answer 2

We can start by rearranging the given equation:12 cos(theta) - 16 sin(theta) = 0Dividing both sides by 4, we get:3 cos(theta) - 4 sin(theta) = 0Now, we can use the trigonometric identity sin^2(theta) + cos^2(theta) = 1 to rewrite sin(theta) in terms of cos(theta):sin(theta) = sqrt(1 - cos^2(theta))Substituting this into the rearranged equation, we get:3 cos(theta) - 4 sqrt(1 - cos^2(theta)) = 0Squaring both sides, we get:9 cos^2(theta) - 16 (1 - cos^2(theta)) = 0Simplifying, we get:25 cos^2(theta) = 16cos(theta) = ±4/5If cos(theta) = 4/5, then sin(theta) = 3/5 (from the identity sin^2(theta) + cos^2(theta) = 1)So, 2 sin(theta) + cos(theta) = 2(3/5) + 4/5 = 6/5 + 4/5 = 2If cos(theta) = -4/5, then sin(theta) = -3/5So, 2 sin(theta) + cos(theta) = 2(-3/5) - 4/5 = -6/5 - 4/5 = -2Therefore, the two possible values of 2 sin(theta) + cos(theta) are 2 and -2.

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