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

The general solution for 4cos^2(x)-2sin(x)+1=0 is x=1.10987…+2pin,x=pi-1.10987…+2pin

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

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

x = 1.10987 \dots + 2 πn, x = π - 1.10987 \dots + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = - \frac{1 + 21}{4}, sin (x) = \frac{21 - 1}{4}

sin (x) = - \frac{1 + 21}{4} :

No Solution

sin (x) = \frac{21 - 1}{4} : x = arcsin (\frac{21 - 1}{4}) + 2 πn, x = π - arcsin (\frac{21 - 1}{4}) + 2 πn

Combine all the solutions

x = arcsin (\frac{21 - 1}{4}) + 2 πn, x = π - arcsin (\frac{21 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 1.10987 \dots + 2 πn, x = π - 1.10987 \dots + 2 πn

tan (x) = \frac{3 3}{5}

tan (x) + 1 = - \frac{1}{3} - \frac{1}{3} cot (x)

6 cos^{2} (x) + 7 cos (x) + 2 = 0

sin (θ) = \frac{4}{5}, 0 < θ < \frac{π}{2}, sin (2 θ)

sin (x) - 4 = cos (x) - 4

What is the general solution for 4cos^2(x)-2sin(x)+1=0 ?
The general solution for 4cos^2(x)-2sin(x)+1=0 is x=1.10987…+2pin,x=pi-1.10987…+2pin