What is the general solution for sin(2x)+sin^2(x)+cos^2(x)=0 ?

The general solution for sin(2x)+sin^2(x)+cos^2(x)=0 is x=(3pi)/4+pin

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sin(2x)+sin^2(x)+cos^2(x)=0

sin (2 x) + sin^{2} (x) + cos^{2} (x) = 0

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 x) + sin^{2} (x) + cos^{2} (x) = 0

Rewrite using trig identities

sin (2 x) + sin^{2} (x) + cos^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= sin (2 x) + 1

sin (2 x) + 1 = 0

Move

1

to the right side

sin (2 x) + 1 = 0

Subtract

1

from both sides

sin (2 x) + 1 - 1 = 0 - 1

Simplify

sin (2 x) = - 1

sin (2 x) = - 1

General solutions for

sin (2 x) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

Solve

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

csc (x) = \frac{1}{cos ( x )}

csc (θ) = (1.201)

3 cos (θ) + 2 = 5 sec (θ), 0 \leq θ < 2 π

2 cos (\frac{π}{4} - 3 x) = 2

3 sec^{2} (x) = sec (x)

What is the general solution for sin(2x)+sin^2(x)+cos^2(x)=0 ?
The general solution for sin(2x)+sin^2(x)+cos^2(x)=0 is x=(3pi)/4+pin