What is the general solution for sin(x+20)+sin(40-x)=1 ?

The general solution for sin(x+20)+sin(40-x)=1 is x=360n+10

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sin(x+20)+sin(40-x)=1

sin (x + 2 0^{\circ}) + sin (4 0^{\circ} - x) = 1

x = 36 0^{\circ} n + 1 0^{\circ}

Solution steps

sin (x + 2 0^{\circ}) + sin (4 0^{\circ} - x) = 1

Rewrite using trig identities

cos (\frac{18 x - 18 0 ^{\circ}}{18}) = 1

General solutions for

cos (\frac{18 x - 18 0 ^{\circ}}{18}) = 1

\frac{18 x - 18 0 ^{\circ}}{18} = 0 + 36 0^{\circ} n

Solve

\frac{18 x - 18 0 ^{\circ}}{18} = 0 + 36 0^{\circ} n : x = 36 0^{\circ} n + 1 0^{\circ}

x = 36 0^{\circ} n + 1 0^{\circ}

x, f = 2 cos (3 x^{2} - 1) e n t o n ces f

tan (a) = \frac{5}{8}

x, z = tan (\frac{x}{2})

tan (\frac{x}{2}) = 3, 0 \leq x \leq 2 π

0 = 4 sin (2 θ)

What is the general solution for sin(x+20)+sin(40-x)=1 ?
The general solution for sin(x+20)+sin(40-x)=1 is x=360n+10