What is the general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi ?

The general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi is x=0,x=2pi

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sin^2(x)-9cos(x)+9=0,0<= x<= 2pi

sin^{2} (x) - 9 cos (x) + 9 = 0, 0 \leq x \leq 2 π

x = 0, x = 2 π

Solution steps

sin^{2} (x) - 9 cos (x) + 9 = 0, 0 \leq x \leq 2 π

Rewrite using trig identities

10 - cos^{2} (x) - 9 cos (x) = 0

Solve by substitution

cos (x) = - 10, cos (x) = 1

cos (x) = - 10, 0 \leq x \leq 2 π :

No Solution

cos (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

Combine all the solutions

x = 0, x = 2 π

- 10 cos (2 x) - 32 cos (x) - 22 = 0

θ, sin (θ) = - \frac{3}{4}

tan (x) \cdot sec (x) = 2

cos (2 x + \frac{π}{3}) = sin (2 x)

cot (2 t) = 0

What is the general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi ?
The general solution for sin^2(x)-9cos(x)+9=0,0<= x<= 2pi is x=0,x=2pi