How do you solve this math problem involving sin and cos?

Question

How do you solve this?

47+65sinx=56cos^2x, where 0 degrees
The answer is 7.18 degrees and 172.82 degrees...but how do you DO IT??? I can use a scientific calculator......can I use that....but first don't i need to simply the equation first or something?

Answer 1

47 + 65sinx = 56(cosx)^2

cosx^2 = 1 - sinx^2

47 + 65sinx = 56(1 - sinx^2)
47 + 65sinx = 56 - 56sinx^2
56sinx^2 + 65sinx - 9 = 0

think of this like

56x^2 + 65x - 9 = 0

x = (-b ± sqrt(b^2 - 4ac))/2a

x = (-65 ± sqrt((65^2 - 4(56)(-9)))/(2(56))
x = (-65 ± sqrt(4225 + 2016))/112
x = (-65 ± sqrt(6241))/112
x = (-65 ± 79)/112
x = (-144/112) or (14/112)
x = (-9/7) or (1/8)

so this gives you

sinx = (-9/7) or sinx = (1/8)
sin(x) = (-9/7) will become undefined, so

sinx = (1/8)
x = sin^-1(1/8)
x = 7.18075578

if you subtract that from 180, you get

x = 172.819244

Don't ask me how i got that, or why i did it that way, i was just trying to come up with the other degrees i couldn't.

Answer 2

We have:

47+65sinx=56cos^2x

We know that cos^2x+sin^2x=1, so cos^2x=1-sin^2x

So now 47+65sinx=56-56sin^2x

56sin^2x+65sinx-9=0

Let X=sinx
Then: 56X^2+65X-9=0, a quadratic!

Solving with the general quadratic formula, the discriminant is equal to 6241=79^2

So X=(-65+79)/2*56 or X=(-65-79)/2*56
X=14/112 or X=144/112.

You can exclude X=144/112, because 144/112>1 and yet we must have -1
So the equation becomes: sinx=14/112. Take the arcsine of 14/112 and you get:

x=7.18. As you know, supplementary angles have the same sin, so x=180-7.18=172.82 degrees is also a solution.

Conclusion: x=7.18 degrees or x=172.82 degrees

Answer 3

you need to know your special triangles: 30-60-90 and 45-45-90 pi/4 = 45 degrees pi/6 = 30 degrees pi/3 = 60 degrees the triangle (and angle) you choose will be based on the acute angle the given angle makes with the x-axis 1) 7pi/4 is the same as -pi/4, or a 45-45-90 triangle in Quadrant 4, where sin is negative and cos is positive sin = y / r = -sqrt(2) / 2 2) cos (-3pi/4) is a 45-45-90 degree triangle in Quadrant 3, where cos and sin are both negative cos (-3pi/4) = -sqrt(2) / 2 3) 7pi/6 is a 30-60-90 triangle in Quadrant 3, with the 30 degree angle at the origin. Since cos and sin are both negative, their quotient (tan) is positive y = -1/2 and x = -sqrt(3) / 2 tan (7pi/6)= y/x = (-1/2) / (-sqrt(3) / 2) = 1 / sqrt(3) = sqrt(3) / 3 it will make your life MUCH easier in trig if you know these special triangles, and their places on the unit circle typically, when looking for the trig values of these angles, you're looking for exact values (sqrt(2) / 2 , not .707; sqrt(3) / 2, not .866)

Answer 4

I don't have a calculator with me right now, but I would suggest using the trig identity cos^2(x) = (1-sin^2(x)), and moving all the expressions to one side so you have a quadratic equation that you can solve using the quadratic formula.

Answer 5

(cos x)^2 = 1 - (sin x)^2

47 + 65 sin x = 56 (1- (sin x)^2)
47 + 65 sin x = 56 - 56 (sin x)^2
56 (sin x)^2 + 65 sin x - 9 = 0

using the quadratic formula, sin x = 0.125
so, x = 7.18

Answer 6

cos2x = (cosx)^2-(sinx)^2 = 2(cosx)^2-1 = 1-2(sinx)^2

47 + 65 sin x = 56 - 112 (sin x) ^2

=> 112(sin x)^2 +65 sin x -9 = 0

Let z = sin x

=> 112z^2 + 65z - 9 = 0

Use quadratic formula from here.

Answer 7

You can use the sum formulas for cosine and sine....
cos2x=2sinxcosx to get rid of the 2x. My advice would be just to graph it and find the x intercepts.

Answer 8

Oh man.. I am glad my kids are in elementry school.. I dread the day I have to help with this! Did you write that in english???? lol... I am scared!

Answer 9

damn, this is pathetic, i just learned this awhile ago, but i cna't remember how to do this.

Answer 10

do you have the rules?