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

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

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

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

x = πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

1 = cos (2 x)

Switch sides

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

(218) sin^{2} (x) + (126) sin (x) - 88 = 0

3 sin (3 x) = 0

cos (θ) csc (θ) = sin (θ) sec (θ)

64 cos^{2} (θ) = 49, (0^{\circ} \leq, θ \leq 36 0^{\circ})

y, x = 4 sec (y)

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