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

The general solution for -2sin(x)+cos(2x)=0 is x=0.37473…+2pin,x=pi-0.37473…+2pin

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-2sin(x)+cos(2x)=0

- 2 sin (x) + cos (2 x) = 0

x = 0.37473 \dots + 2 πn, x = π - 0.37473 \dots + 2 πn

Solution steps

- 2 sin (x) + cos (2 x) = 0

Rewrite using trig identities

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

Solve by substitution

sin (x) = - \frac{1 + 3}{2}, sin (x) = \frac{3 - 1}{2}

sin (x) = - \frac{1 + 3}{2} :

No Solution

sin (x) = \frac{3 - 1}{2} : x = arcsin (\frac{3 - 1}{2}) + 2 πn, x = π - arcsin (\frac{3 - 1}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{3 - 1}{2}) + 2 πn, x = π - arcsin (\frac{3 - 1}{2}) + 2 πn

Show solutions in decimal form

x = 0.37473 \dots + 2 πn, x = π - 0.37473 \dots + 2 πn

X, 4 cos (X) cos (Y) = 3

0 = 4 sin (\frac{π}{2} (x - 2)) + 2

sin (θ) = \frac{3}{8}, sin (2 0^{\circ})

- 2 sin (x) + 2 cos (x) = 0

cos (3 x) cos (9 x) + sin (3 x) sin (9 x) = 0, (0, 2 π)

What is the general solution for -2sin(x)+cos(2x)=0 ?
The general solution for -2sin(x)+cos(2x)=0 is x=0.37473…+2pin,x=pi-0.37473…+2pin