What is the general solution for sin(x)+sin^2(x)+cos^2(x)=1 ?

The general solution for sin(x)+sin^2(x)+cos^2(x)=1 is x=2pin,x=pi+2pin

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

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

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= sin (x) + 1

sin (x) + 1 = 1

Subtract

1

from both sides

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

Simplify

sin (x) = 0

General solutions for

sin (x) = 0

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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

tan (3 x) = 5 tan (x)

2 cos (2 x) - 2 = 0

3 tan^{2} (x) + 4 sec (x) = - 4

tan (x + 2 0^{\circ}) \cdot tan (x - 2 0^{\circ}) = 1

18 cos (10 x) + 2 = 10

What is the general solution for sin(x)+sin^2(x)+cos^2(x)=1 ?
The general solution for sin(x)+sin^2(x)+cos^2(x)=1 is x=2pin,x=pi+2pin