What is the general solution for 10sin(x)+10cos^2(x)=10,0<= x<2pi ?

The general solution for 10sin(x)+10cos^2(x)=10,0<= x<2pi is x=0,x=pi,x= pi/2

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10sin(x)+10cos^2(x)=10,0<= x<2pi

10 sin (x) + 10 cos^{2} (x) = 10, 0 \leq x < 2 π

x = 0, x = π, x = \frac{π}{2}

Solution steps

10 sin (x) + 10 cos^{2} (x) = 10, 0 \leq x < 2 π

Subtract

10

from both sides

10 sin (x) + 10 cos^{2} (x) - 10 = 0

Rewrite using trig identities

10 sin (x) - 10 sin^{2} (x) = 0

Solve by substitution

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

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

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

Combine all the solutions

x = 0, x = π, x = \frac{π}{2}

sin (x) - 3 - 3 sin^{2} (x) = 0

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

4 tan^{2} (x) + 21 tan (x) - 49 = 0

cot (x) = csc (x) + 1

\frac{sin ( 8 )}{30} = \frac{sin ( x )}{120}

What is the general solution for 10sin(x)+10cos^2(x)=10,0<= x<2pi ?
The general solution for 10sin(x)+10cos^2(x)=10,0<= x<2pi is x=0,x=pi,x= pi/2