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

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

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

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

x = πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

= cos (2 x)

cos (2 x) = 1

General solutions for

cos (2 x) = 1

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

tan^{2} (x) - 4 sec (x) = 4

sin (x + 1 0^{\circ}) = - tan (1 5^{\circ})

tan (2 θ) = \frac{73 - 8}{- 10}

x, T (7) = 3.15 cos (\frac{π}{6} x) + 19.15

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

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