What is the general solution for 4sin^2(x)+cos(x)+1=0 ?

The general solution for 4sin^2(x)+cos(x)+1=0 is x=pi+2pin

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4sin^2(x)+cos(x)+1=0

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

x = π + 2 πn

Solution steps

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

Rewrite using trig identities

5 + cos (x) - 4 cos^{2} (x) = 0

Solve by substitution

cos (x) = - 1, cos (x) = \frac{5}{4}

cos (x) = - 1 : x = π + 2 πn

cos (x) = \frac{5}{4} :

No Solution

Combine all the solutions

x = π + 2 πn

(cot^{2} (x)) - csc (x) = 1

x, u = arctan (\frac{x}{y})

4 cos^{2} (2 x - 1) = 1

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

3 \cdot sin (x) = cos (x)

What is the general solution for 4sin^2(x)+cos(x)+1=0 ?
The general solution for 4sin^2(x)+cos(x)+1=0 is x=pi+2pin