Question

3sin2x-5sinxcosx+8cos2x=2​

Answer 1

hey mate here is your answer.

Consider,

3sin2 X-5sin X cos X + 8cos2X = 2 ⇒ 3sin2 X-5sin X cos X + 3cos2X + 5cos2X = 2 ⇒ 3sin2 X + 3cos2X - 5sin X cos X + 5cos2X = 2

⇒ 3(sin2 X + cos2X) - 5sin X cos X + 5cos2X = 2 ⇒

3 - 5sin X cos X + 5cos2X = 2

⇒ 5cos2X - 5sin X cos X + 1 = 0 Divide by "sin2x"

⇒ (5cos2X/sin2x) - (5sin X cos X)/(sin2x) + (1/sin2x) = 0

⇒ 5cot2X - 5cot X + cosec2X = 0 ⇒ 5cot2X - 5cot X + (1 + cot2X) = 0 ⇒ 6cot2X - 5cot X + 1 = 0

Solve the quadratic equation.

hope it will help you.

plss brainliest my answer.

Answer 2

$\text{Consider,} 3sin^2 X-5sin X cos X + 8cos^2X = 2\\\implies 3sin^2 X-5sin X cos X + 3cos^2X + 5cos^2X = 2\\\implies 3sin^2X+3cos^2X-5sinXcosX+5cos^2X=2\\\implies 3(sin^2 X+cos^2X) - 5sin X cos X+ 5cos^2X = 2\\\implies 3 - 5sin X cos X + 5cos^2X = 2\\\implies 5cos^2X - 5sin X cosX+ 1 = 0$$\text{Divide by } sin^2X\\\implies (5cos^2X/sin^2X) - (5sin X cos X)/(sin^2X)+ (1/sin^2X)= 0\\\implies 5cot^2X - 5cotX+ cosec^2X = 0\\\implies 5cot^2X-5cotX+ (1 + cot^2X) = 0\\\implies 6cot^2X-5cot X+1 = 0\\\text{Now, Solving the quadratic equation.}$$6cot^2X - 5cot X + 1 = 0\\\implies 6cot^2X - 3cotX -2cotX +1 =0\\\implies3cotX(2cotX-1) -1(2cotX-1) = 0\\\implies (2cotX-1)(1cotX-1)=0\\\\cotX = \frac{1}{2}\\\\ cotX = \frac{1}{1}$