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

The general solution for sin^2(t)+2cos(t)=1 is t= pi/2+2pin,t=(3pi)/2+2pin

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

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

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

- cos^{2} (t) + 2 cos (t) = 0

Solve by substitution

cos (t) = 0, cos (t) = 2

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 2 :

No Solution

Combine all the solutions

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

6 tan (θ) - 2 cot (θ) = 0

- 2 sin^{2} (x) = 7 cos (x) + 2

2 sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

cos (x) (2 cos (x) + 1) = 3

3 + cot (x) = 0

What is the general solution for sin^2(t)+2cos(t)=1 ?
The general solution for sin^2(t)+2cos(t)=1 is t= pi/2+2pin,t=(3pi)/2+2pin