What is the general solution for cos(x/2+20)=sin(x) ?

The general solution for cos(x/2+20)=sin(x) is x=(1260+6480n}{27},x=\frac{1980+6480n)/9

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cos(x/2+20)=sin(x)

cos (\frac{x}{2} + 2 0^{\circ}) = sin (x)

x = \frac{126 0 ^{\circ} + 648 0 ^{\circ} n}{27}, x = \frac{198 0 ^{\circ} + 648 0 ^{\circ} n}{9}

Solution steps

cos (\frac{x}{2} + 2 0^{\circ}) = sin (x)

Rewrite using trig identities

cos (\frac{x}{2} + 2 0^{\circ}) = sin (9 0^{\circ} - (\frac{x}{2} + 2 0^{\circ}))

Apply trig inverse properties

x = 9 0^{\circ} - (\frac{x}{2} + 2 0^{\circ}) + 36 0^{\circ} n, x = 18 0^{\circ} - (9 0^{\circ} - (\frac{x}{2} + 2 0^{\circ})) + 36 0^{\circ} n

x = 9 0^{\circ} - (\frac{x}{2} + 2 0^{\circ}) + 36 0^{\circ} n : x = \frac{126 0 ^{\circ} + 648 0 ^{\circ} n}{27}

x = 18 0^{\circ} - (9 0^{\circ} - (\frac{x}{2} + 2 0^{\circ})) + 36 0^{\circ} n : x = \frac{198 0 ^{\circ} + 648 0 ^{\circ} n}{9}

x = \frac{126 0 ^{\circ} + 648 0 ^{\circ} n}{27}, x = \frac{198 0 ^{\circ} + 648 0 ^{\circ} n}{9}

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

sin (θ) = - \frac{7}{25}, sin (2 θ)

cos (θ) = \frac{9}{12}

csc (x) + cot (x) = - 1, 0 \leq x \leq 2 π

sin (x) + cos (x) = \frac{5}{7}, 9 0^{\circ} > x > 0^{\circ}

What is the general solution for cos(x/2+20)=sin(x) ?
The general solution for cos(x/2+20)=sin(x) is x=(1260+6480n}{27},x=\frac{1980+6480n)/9