What is the general solution for 5sin^2(x)+6cos(x)-6=0 ?

The general solution for 5sin^2(x)+6cos(x)-6=0 is x=1.36943…+2pin,x=2pi-1.36943…+2pin,x=2pin

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5sin^2(x)+6cos(x)-6=0

5 sin^{2} (x) + 6 cos (x) - 6 = 0

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = 2 πn

Solution steps

5 sin^{2} (x) + 6 cos (x) - 6 = 0

Rewrite using trig identities

- 1 - 5 cos^{2} (x) + 6 cos (x) = 0

Solve by substitution

cos (x) = \frac{1}{5}, cos (x) = 1

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

cos (x) = 1 : x = 2 πn

Combine all the solutions

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

Show solutions in decimal form

x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn, x = 2 πn

sin^{2} (x) - sin (x) cos (x) - 6 cos^{2} (x) = 0

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

\frac{cos ^{2} ( x ) + 1}{1 + cot ^{2} ( x )} = 1

3 sin^{2} (c) - 7 sin (x) + 2 = 0

1 + cos^{2} (x) = sin^{4} (x)

What is the general solution for 5sin^2(x)+6cos(x)-6=0 ?
The general solution for 5sin^2(x)+6cos(x)-6=0 is x=1.36943…+2pin,x=2pi-1.36943…+2pin,x=2pin