# Solve equation?

1) sin^3(x)+1/8 =0 if (0≤x≤2pi)

2) -4sin^2(x)+4cos(x)+5=0

3) e^cos(x)=1 if x E (-inf, inf)

### 2 Answers

- Jeff AaronLv 71 month ago
1.

sin^3(x) = -1/8

Since x is real, then sin(x) is real, so we have:

sin(x) = cbrt(-1/8)

sin(x) = -1/2

x = (7/6)*pi radians or x = (11/6)*pi radians

https://www.wolframalpha.com/input/?i=sin%5E3%28x%...

2.

-4*sin^2(x) + 4*cos(x) + 5 = 0

By Pythagoras, sin^2(x) + cos^2(x) = 1, so sin^2(x) = 1 - cos^2(x), so we have:

-4(1 - cos^2(x)) + 4*cos(x) + 5 = 0

-4 + 4*cos^2(x) + 4*cos(x) + 5 = 0

4*cos^2(x) + 4*cos(x) + 1 = 0

(2*cos(x) + 1)^2 = 0

2*cos(x) + 1 = 0

2*cos(x) = -1

cos(x) = -1/2

x = 2*pi*(n +/- (1/3)) radians, for any integer n

https://www.wolframalpha.com/input/?i=-4*sin%5E2%2...

3.

e^cos(x) = 1

Since x is a real number, then cos(x) is a real number, so we have:

cos(x) = ln(1)

cos(x) = 0

x = pi*(n - 0.5) radians, for any integer n

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- nyphdinmdLv 71 month ago
1. Sin^3(x) + 1/8 = 0 --> sin^3(x) = -1/8 take cube root --> sin(x) = -1/2, +1/2

Note the +1/2 root is degenerate - it can appear twice with a -1/2 root and the the -1/2 root is triple degenerate i.e. (-1/2)^3 = - (1/2)^3 = -1/8

for the positive root x = pi/4, 3*pi/4

for the negative root x = 5*pi/4, 7*pi/4

2. -4sin^2(x) + 4 cos(x) + 5 = 0 --> use sin^2(x) = 1 - cos^2(x)

-4 + 4cos^2(x) + 4cos(x) + 5 = 0 --> 4cos^2(x) + 4cos(x) + 1 = 0

let cos(x) = u --> 4u^2 + 4u +1 = 0 --> u = -2 +/- (1/2)sqrt(16-4) = -2 +/- sqrt(3)

since u = cos(x) -1 <= u < = 1 so only +sqrt(3) satisfies this condition

u = -0.2679 --> cos(x) = -0.2679 x = 1.842 radians

3. e^(cos(x)) = 1

take ln of both sides --> cos(x) = 0 --> x = m*pi/2 where m = all odd negative and positive integers

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