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

The general solution for cos(2x)+sin(x)=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x= pi/2

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cos(2x)+sin(x)=0,0<= x<= 2pi

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

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{2}

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = - \frac{1}{2}, sin (x) = 1

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{2}

2 sin (θ) + 5 = 6

x, cos (x) = \frac{1}{2}

2 sin (2 x) + 3 = 0

20 sin (x) + 3 = 8

cos (2 θ) + 3 = 5 cos (θ)

What is the general solution for cos(2x)+sin(x)=0,0<= x<= 2pi ?
The general solution for cos(2x)+sin(x)=0,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x= pi/2