What is the general solution for -5sin(x)=-2cos^2(x)+4,0<= x<= 2pi ?

The general solution for -5sin(x)=-2cos^2(x)+4,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6

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-5sin(x)=-2cos^2(x)+4,0<= x<= 2pi

- 5 sin (x) = - 2 cos^{2} (x) + 4, 0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Solution steps

- 5 sin (x) = - 2 cos^{2} (x) + 4, 0 \leq x \leq 2 π

Subtract

- 2 cos^{2} (x) + 4

from both sides

- 5 sin (x) + 2 cos^{2} (x) - 4 = 0

Rewrite using trig identities

- 2 - 2 sin^{2} (x) - 5 sin (x) = 0

Solve by substitution

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

sin (x) = - 2, 0 \leq x \leq 2 π :

No Solution

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

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}

6 = sin (\frac{π}{4} x) + 5

2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

t, 4.6 = 0.106 cos (4.36 t)

sin (x) = \frac{23}{24}

5 - 5 cos (x) = 4 sin^{2} (x)

What is the general solution for -5sin(x)=-2cos^2(x)+4,0<= x<= 2pi ?
The general solution for -5sin(x)=-2cos^2(x)+4,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6