What is the general solution for sin(3x+10)=cos(x+24) ?

The general solution for sin(3x+10)=cos(x+24) is x=(10800n+1980-300)/(120),x=(3420+10800n-300)/(60)

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sin(3x+10)=cos(x+24)

sin (3 x + 10) = cos (x + 2 4^{\circ})

x = \frac{1080 0 ^{\circ} n + 198 0 ^{\circ} - 300}{120}, x = \frac{342 0 ^{\circ} + 1080 0 ^{\circ} n - 300}{60}

Solution steps

sin (3 x + 10) = cos (x + 2 4^{\circ})

Rewrite using trig identities

sin (3 x + 10) = sin (9 0^{\circ} - (x + 2 4^{\circ}))

Apply trig inverse properties

3 x + 10 = 9 0^{\circ} - (x + 2 4^{\circ}) + 36 0^{\circ} n, 3 x + 10 = 18 0^{\circ} - (9 0^{\circ} - (x + 2 4^{\circ})) + 36 0^{\circ} n

3 x + 10 = 9 0^{\circ} - (x + 2 4^{\circ}) + 36 0^{\circ} n : x = \frac{1080 0 ^{\circ} n + 198 0 ^{\circ} - 300}{120}

3 x + 10 = 18 0^{\circ} - (9 0^{\circ} - (x + 2 4^{\circ})) + 36 0^{\circ} n : x = \frac{342 0 ^{\circ} + 1080 0 ^{\circ} n - 300}{60}

x = \frac{1080 0 ^{\circ} n + 198 0 ^{\circ} - 300}{120}, x = \frac{342 0 ^{\circ} + 1080 0 ^{\circ} n - 300}{60}

tan (x^{2}) + 1 = 0

\frac{( 1 + cos ^{2} ( a ) )}{sin ^{2} ( a )} = \frac{5}{3}

d^{2} + 13 d + 36 = \frac{sin ^{2} ( x )}{2}

\frac{tan ^{2} ( x ) - 4}{cos ( x ) + 5} = 0

cot^{2} (x) = sec^{2} (x) - 1

What is the general solution for sin(3x+10)=cos(x+24) ?
The general solution for sin(3x+10)=cos(x+24) is x=(10800n+1980-300)/(120),x=(3420+10800n-300)/(60)