What is the general solution for cos^2(x)=tan(x) ?

The general solution for cos^2(x)=tan(x) is x=0.59876…+2pin,x=-2.54282…+2pin

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cos^2(x)=tan(x)

cos^{2} (x) = tan (x)

x = 0.59876 \dots + 2 πn, x = - 2.54282 \dots + 2 πn

Solution steps

cos^{2} (x) = tan (x)

Subtract

tan (x)

from both sides

cos^{2} (x) - tan (x) = 0

Express with sin, cos

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

Simplify

cos^{2} (x) - \frac{sin ( x )}{cos ( x )} : \frac{cos ^{3} ( x ) - sin ( x )}{cos ( x )}

\frac{cos ^{3} ( x ) - sin ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Add

sin (x)

to both sides

cos^{3} (x) = sin (x)

Square both sides

(cos^{3} (x))^{2} = sin^{2} (x)

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = 0.68232 \dots, cos (x) = - 0.68232 \dots

cos (x) = 0.68232 \dots : x = arccos (0.68232 \dots) + 2 πn, x = 2 π - arccos (0.68232 \dots) + 2 πn

cos (x) = - 0.68232 \dots : x = arccos (- 0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

Combine all the solutions

x = arccos (0.68232 \dots) + 2 πn, x = 2 π - arccos (0.68232 \dots) + 2 πn, x = arccos (- 0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

Verify solutions by plugging them into the original equation

x = arccos (0.68232 \dots) + 2 πn, x = - arccos (- 0.68232 \dots) + 2 πn

Show solutions in decimal form

x = 0.59876 \dots + 2 πn, x = - 2.54282 \dots + 2 πn

sin^{2} (x) - 6 sin (x) + 3 = 0

12 csc^{2} (θ) - csc (θ) - 1 = 0

2 cos (2 θ) - 4 cos (θ) + 2 = - 1

cos (a) = \frac{1}{sin ( a )}

sin (a) = cos (a)

What is the general solution for cos^2(x)=tan(x) ?
The general solution for cos^2(x)=tan(x) is x=0.59876…+2pin,x=-2.54282…+2pin