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

The general solution for 4cos(x)=sin^2(x)+1 is x=1.10460…+2pin,x=2pi-1.10460…+2pin

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

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

x = 1.10460 \dots + 2 πn, x = 2 π - 1.10460 \dots + 2 πn

Solution steps

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

Subtract

sin^{2} (x) + 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = - 2 + 6, cos (x) = - 2 - 6

cos (x) = - 2 + 6 : x = arccos (- 2 + 6) + 2 πn, x = 2 π - arccos (- 2 + 6) + 2 πn

cos (x) = - 2 - 6 :

No Solution

Combine all the solutions

x = arccos (- 2 + 6) + 2 πn, x = 2 π - arccos (- 2 + 6) + 2 πn

Show solutions in decimal form

x = 1.10460 \dots + 2 πn, x = 2 π - 1.10460 \dots + 2 πn

cos^{2} (x) + sin^{2} (x) - 2 sin (x) = 0, 0 \leq x \leq 2 π

cot (A) = \frac{24}{7}

23 sec (θ) + 4 = 0

3 sin (x) = 2

\frac{sin ( 7 4 ^{\circ} )}{8} = \frac{sin ( a )}{4}

What is the general solution for 4cos(x)=sin^2(x)+1 ?
The general solution for 4cos(x)=sin^2(x)+1 is x=1.10460…+2pin,x=2pi-1.10460…+2pin