What is the general solution for sin(2x+60)+sin(x+30)=0,0<= x<= 2pi ?

The general solution for sin(2x+60)+sin(x+30)=0,0<= x<= 2pi is x=11pi-30

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sin(2x+60)+sin(x+30)=0,0<= x<= 2pi

sin (2 x + 60) + sin (x + 30) = 0, 0 \leq x \leq 2 π

x = 11 π - 30

Solution steps

sin (2 x + 60) + sin (x + 30) = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

2 cos (\frac{- 30 - x}{2}) sin (\frac{90 + 3 x}{2}) = 0

Solving each part separately

cos (\frac{- 30 - x}{2}) = 0 or sin (\frac{90 + 3 x}{2}) = 0

cos (\frac{- 30 - x}{2}) = 0, 0 \leq x \leq 2 π : x = 11 π - 30

sin (\frac{90 + 3 x}{2}) = 0, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = 11 π - 30

tan^{2} (x) = 3 - csc^{2} (x)

2 sin^{2} (u) = 1 + sin (u)

2 tan^{2} (θ) - 3 tan (θ) + 1 = 0

cos (x) = \frac{9}{17}

- 6 sin (c) + 0 = sin (c) - 3

What is the general solution for sin(2x+60)+sin(x+30)=0,0<= x<= 2pi ?
The general solution for sin(2x+60)+sin(x+30)=0,0<= x<= 2pi is x=11pi-30