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

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

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

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

x = 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

cos (2 x) cos (x) + sin (2 x) sin (x)

Use the Angle Difference identity:

cos (s) cos (t) + sin (s) sin (t) = cos (s - t)

= cos (2 x - x)

cos (2 x - x) = 1

General solutions for

cos (2 x - 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 - x = 0 + 2 πn

2 x - x = 0 + 2 πn

Solve

2 x - x = 0 + 2 πn : x = 2 πn

2 x - x = 0 + 2 πn

Add similar elements:

2 x - x = x

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (2 θ) + 10 sin^{2} (θ) = 7

4 cos (2 t) + 1 = 3

cos (2 x) - cos (x) = - 1

cos (x + \frac{π}{4}) - cos (x - \frac{π}{4}) = 1

y, x = cos (y)

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