What is the general solution for cos(x)cos(35)=1-sin(x)sin(35) ?

The general solution for cos(x)cos(35)=1-sin(x)sin(35) is x=-360n+35

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cos(x)cos(35)=1-sin(x)sin(35)

cos (x) cos (3 5^{\circ}) = 1 - sin (x) sin (3 5^{\circ})

x = - 36 0^{\circ} n + 3 5^{\circ}

Solution steps

cos (x) cos (3 5^{\circ}) = 1 - sin (x) sin (3 5^{\circ})

Subtract

1 - sin (x) sin (3 5^{\circ})

from both sides

cos (3 5^{\circ}) cos (x) - 1 + sin (3 5^{\circ}) sin (x) = 0

Rewrite using trig identities

- 1 + cos (3 5^{\circ} - x) = 0

Move

1

to the right side

cos (3 5^{\circ} - x) = 1

General solutions for

cos (3 5^{\circ} - x) = 1

3 5^{\circ} - x = 0 + 36 0^{\circ} n

Solve

3 5^{\circ} - x = 0 + 36 0^{\circ} n : x = - 36 0^{\circ} n + 3 5^{\circ}

x = - 36 0^{\circ} n + 3 5^{\circ}

4 sin^{2} (x) + 2 sin (x) = 2, 0 \leq x \leq 2 π

3 tan (\frac{1}{3} x) = 8

1.6 5^{2} = 0.6 5^{2} + 1.7 7^{2} - 2 (0.65) (1.77) cos (A)

x, cos (3 x + \frac{π}{4}) = \frac{1}{2}

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

What is the general solution for cos(x)cos(35)=1-sin(x)sin(35) ?
The general solution for cos(x)cos(35)=1-sin(x)sin(35) is x=-360n+35