What is the general solution for cos^2(3x)-sin^2(3x)=-1 ?

The general solution for cos^2(3x)-sin^2(3x)=-1 is x= pi/6+(pin)/3

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cos^2(3x)-sin^2(3x)=-1

cos^{2} (3 x) - sin^{2} (3 x) = - 1

x = \frac{π}{6} + \frac{πn}{3}

Degrees

x = 3 0^{\circ} + 6 0^{\circ} n

Solution steps

cos^{2} (3 x) - sin^{2} (3 x) = - 1

Rewrite using trig identities

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

Use the Double Angle identity:

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

= cos (2 \cdot 3 x)

cos (2 \cdot 3 x) = - 1

General solutions for

cos (2 \cdot 3 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 \cdot 3 x = π + 2 πn

2 \cdot 3 x = π + 2 πn

Solve

2 \cdot 3 x = π + 2 πn : x = \frac{π}{6} + \frac{πn}{3}

2 \cdot 3 x = π + 2 πn

Divide both sides by

6

2 \cdot 3 x = π + 2 πn

Divide both sides by

6

\frac{2 \cdot 3 x}{6} = \frac{π}{6} + \frac{2 πn}{6}

Simplify

x = \frac{π}{6} + \frac{πn}{3}

x = \frac{π}{6} + \frac{πn}{3}

x = \frac{π}{6} + \frac{πn}{3}

x, arctan (\frac{y}{x}) = \frac{π}{4}

(1 + cos (x)) (tan (x) - 1) = 0

cos (x) = \pm \frac{1}{2}

\frac{cos ( θ )}{1 + sin ( θ )} = tan (\frac{π}{4} - θ)

sin (x) = \frac{1}{4 sin ( x )}

What is the general solution for cos^2(3x)-sin^2(3x)=-1 ?
The general solution for cos^2(3x)-sin^2(3x)=-1 is x= pi/6+(pin)/3