help with some math problems, does anyone know how to do this .Appreciate it alot if helped.?

Question

*Solve for X, cos square x - 2sin X = 2

*A=1.5, B=1.7, C=1.8, find the smallest angle in degree.

*if cos of X=4/7 and X is in the fourth quadrant, find the value of the five remaining trig

Answer 1

(cosx)^2 - 2sinx = 2
cosx^2 = 1 - sinx^2
(1 - sinx^2) - 2sinx = 2
1 - sinx^2 - 2sinx = 2
-sinx^2 - 2sinx + 1 = 2
-sinx^2 - 2sinx - 1 = 0
-(sinx^2 + 2sinx + 1) = 0
-(sinx + 1)(sinx + 1)

-sinx - 1 = 0
-sinx = 1
sinx = -1
x = sin^-1(-1)
x = 270°

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1.5^2 = 1.7^2 + 1.8^2 - 2(1.7)(1.8)cosA
2.25 = 2.89 + 3.24 - 6.12cosA
2.25 = 6.13 - 6.12cosA
-6.12cosA = -3.88
cosA = (388/612)
cosA = (97/153)
A = 50.6551127

1.7^2 = 1.5^2 + 1.8^2 - 2(1.5)(1.8)cosB
2.89 = 2.25 + 3.24 - 5.4cosB
2.89 = 5.49 - 5.4cosB
-2.6 = -5.4cosB
cosB = (13/27)
B = 61.2177953

1.8^2 = 1.5^2 + 1.7^2 - 2(1.5)(1.7)cos(C)
3.24 = 2.25 + 2.89 - 5.1cosC
3.24 = 5.14 - 5.1cosC
-1.9 = -5.1cosC
cosC = (19/51)
C = 68.1270919

As you can see, the angle opposite of side A, is the smallest angle.

ANS : about 51°

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cos(A) = 4/7
x = 4
r = 7

r = sqrt(x^2 + y^2)
7 = sqrt(4^2 + y^2)
49 = 16 + y^2
y^2 = 33
y = sqrt(33)

sin(A) = y/r
sinA = (-sqrt(33))/7

tanA = y/x
tanA = (-sqrt(33))/4

cscA = r/y
cscA = -7/(sqrt(33))
cscA = (-7sqrt(33))/33

secA = r/x
secA = 7/4

cotA = x/y
cotA = -4/(sqrt(33))
cotA = (-4sqrt(33))/33

Answer 2

1) Since cos^2(x) + sin^2(x) = 1, we can plug in

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

which gives us:

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

If we substitute

u = sin(x),

then we have a quadratic equation:

-u^2+1-2u=2
or
u^2 + 2u + 1 = 0

You can solve for u by factoring:

(u + 1)(u + 1) = 0 --> u = -1

So

u = -1 ---> sin(x) = -1 ---> x = 3pi/2 (radians) = 270 (degrees)


2) I will not calculate this all the way out, but you should be able to use the law of cosines:

A^2 = B^2 + C^2 - 2BC cos(a)

where A, B, and C are the side lengths, and a is the angle opposite side A. You want to find angle a because the smallest angle is always opposite the shortest side.


3) Knowing that cos(x) = 4/7 allows you to construct a triangle. If we make the hypotenuse of length 7, then the side adjacent to x must be 4, and using the pythagorean theorem, the last side must be Sqrt(49 - 16) = Sqrt(33). Since we are in the fourth quadrant, the last side must be negative, because the opposite side extends down from the x-axis. So we have

Adjacent = 4
Opposite = -Sqrt(33)
Hypotenuse = 7

You should be able to determine all of the trig function values from this information.

I hope this helps; these can be tricky concepts.