sin^2x-8sinx-4=0?Find all values of x in the interval [0, 360deg.] that satisfy the equation. Round approximat?

Question

Find all values of x in the interval [0, 360deg.] that satisfy the equation. Round approximate answers to the nearest tenth of a degree. Please show the work. sin^2x-8sinx-4=0

Answer 1

sin^2x-8sinx-4=0

sinx = [8±√(8²+4*4)]/2

sinx = (8±√80)/2 = 4±2√5 = -0.47214 (discard the positive value as outside the range of the sine function)



x = asin(-0.47214) = 208.2°, 331.8°

Answer 2

The gentleman above is right. I'm however eager to go through the steps.



To avoid confusion, let sin x = y



So y^2 - 8y - 4 = 0. This cannot be factorised so I'm going to complete the squares.



How? Half the coefficient of y, square it and add and subtract it. What??



See. y^2 - 8y + (-8/2)^2 - (-8/2)^2 - 4 = 0



y^2 - 8y + 16 - 16 - 4 = 0. Think about this +16 and -16 do not add any thing to the equation, they are just there to help me factorize, to make life easy!



So (y - 4)^2 = 20



y - 4 = +/- 2sqrt5



y = 4+2sqrt5 = 8.47 or y = 4-2sqrt5 = -0.47



But remember that y = sin x, so sin x = 8.47 (no solution)



sin x = -0.47 sine is negative on the 3rd and 4th quadrant of a unit circle. Hope you realise this.



Any way what you gotta do is find arc sin or sin^-1 0.47 the normal way to get 28 degrees, then you add on 180, for it to fall on the 3rd quadrant 180 + 28 = 208 degrees.

You subtract 28 from 360, for it to fall on the 4th quadrant, 360 - 28 = 332 degrees



CHECK.

sin^2 208 - 8sin 208 - 4 = 3.98 - 4 = approximately equal so correct.



Same goes for 332 degrees.



Hope it's clear.



Good luck

Answer 3

substitute sinx=y

y^2-8y-4=0

quadratic formula will give you two answers

use inverse sin to find the angles

Answer 4

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