What is the general solution for cos^2(x)+sin(x)=1,0<= x<= 2pi ?

The general solution for cos^2(x)+sin(x)=1,0<= x<= 2pi is x=0,x=pi,x=2pi,x= pi/2

Popular >

cos^2(x)+sin(x)=1,0<= x<= 2pi

cos^{2} (x) + sin (x) = 1, 0 \leq x \leq 2 π

x = 0, x = π, x = 2 π, x = \frac{π}{2}

Solution steps

cos^{2} (x) + sin (x) = 1, 0 \leq x \leq 2 π

Subtract

1

from both sides

cos^{2} (x) + sin (x) - 1 = 0

Rewrite using trig identities

sin (x) - sin^{2} (x) = 0

Solve by substitution

sin (x) = 0, sin (x) = 1

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

Combine all the solutions

x = 0, x = π, x = 2 π, x = \frac{π}{2}

1 + tan^{2} (x) = cot^{2} (x)

1 - 4 tan (3 x + 1) = 17

sin (x) cos (x) + sin (x) = 3 cos^{2} (x) + 3 cos (x)

2 - 2 sin^{2} (x) = - 3 cos (x) + 2

4 tan (θ) + 7 = 0

What is the general solution for cos^2(x)+sin(x)=1,0<= x<= 2pi ?
The general solution for cos^2(x)+sin(x)=1,0<= x<= 2pi is x=0,x=pi,x=2pi,x= pi/2