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

The general solution for sin^2(x)-sin(x)+2cos^2(x)=1 is x=0.66623…+2pin,x=pi-0.66623…+2pin

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

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

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2} :

No Solution

sin (x) = \frac{5 - 1}{2} : x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

Show solutions in decimal form

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

\frac{sin ^{3} ( x )}{2 + 2 ( sin ( x ) ) ^{2}} = 1

x, cot^{2} (x) = \frac{1}{3}

2 + cos^{2} (x) = - 5 sin (x)

tan^{5} (x) tan^{2} (x) = 1

\frac{sin ^{2} ( a ) + 1}{( 1 + tan ^{2} ( a ) )} = 1

What is the general solution for sin^2(x)-sin(x)+2cos^2(x)=1 ?
The general solution for sin^2(x)-sin(x)+2cos^2(x)=1 is x=0.66623…+2pin,x=pi-0.66623…+2pin