What is the general solution for cos^2(x)=(1-tan^2(x))/(sec^2(x)) ?

The general solution for cos^2(x)=(1-tan^2(x))/(sec^2(x)) is x=pin

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cos^2(x)=(1-tan^2(x))/(sec^2(x))

cos^{2} (x) = \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )}

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

cos^{2} (x) = \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )}

Subtract

\frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )}

from both sides

cos^{2} (x) - \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )} = 0

Simplify

cos^{2} (x) - \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )} : \frac{cos ^{2} ( x ) sec ^{2} ( x ) - 1 + tan ^{2} ( x )}{sec ^{2} ( x )}

cos^{2} (x) - \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )}

Convert element to fraction:

cos^{2} (x) = \frac{cos ^{2} ( x ) sec ^{2} ( x )}{sec ^{2} ( x )}

= \frac{cos ^{2} ( x ) sec ^{2} ( x )}{sec ^{2} ( x )} - \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cos ^{2} ( x ) sec ^{2} ( x ) - ( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )}

- (1 - tan^{2} (x)) : - 1 + tan^{2} (x)

- (1 - tan^{2} (x))

Distribute parentheses

= - (1) - (- tan^{2} (x))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + tan^{2} (x)

= \frac{cos ^{2} ( x ) sec ^{2} ( x ) - 1 + tan ^{2} ( x )}{sec ^{2} ( x )}

\frac{cos ^{2} ( x ) sec ^{2} ( x ) - 1 + tan ^{2} ( x )}{sec ^{2} ( x )} = 0

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

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

Rewrite using trig identities

- 1 + tan^{2} (x) + cos^{2} (x) sec^{2} (x)

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

= - 1 + tan^{2} (x) + (\frac{1}{sec ( x )})^{2} sec^{2} (x)

Simplify

- 1 + tan^{2} (x) + (\frac{1}{sec ( x )})^{2} sec^{2} (x) : tan^{2} (x)

- 1 + tan^{2} (x) + (\frac{1}{sec ( x )})^{2} sec^{2} (x)

(\frac{1}{sec ( x )})^{2} sec^{2} (x) = 1

(\frac{1}{sec ( x )})^{2} sec^{2} (x)

(\frac{1}{sec ( x )})^{2} = \frac{1}{sec ^{2} ( x )}

(\frac{1}{sec ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sec ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( x )}

= \frac{1}{sec ^{2} ( x )} sec^{2} (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sec ^{2} ( x )}{sec ^{2} ( x )}

Cancel the common factor:

sec^{2} (x)

= 1

= - 1 + tan^{2} (x) + 1

Group like terms

= tan^{2} (x) - 1 + 1

- 1 + 1 = 0

= tan^{2} (x)

= tan^{2} (x)

tan^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

tan (x) = 0

General solutions for

tan (x) = 0

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

sin^{2} (x) - 3 cos (x) - 4 = 0

tan^{2} (x) + sec (x) = 5

4 sin^{5} (x) = 3

cos (\frac{x}{2}) = 0.5

sin (x + 6 0^{\circ}) = 2 cos (x)

What is the general solution for cos^2(x)=(1-tan^2(x))/(sec^2(x)) ?
The general solution for cos^2(x)=(1-tan^2(x))/(sec^2(x)) is x=pin