What is the general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 ?

The general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 is x=210,x=330,x=0.84806…,x=180-0.84806…

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4cos(2x)+2sin(x)=1,0<= x<= 360

4 cos (2 x) + 2 sin (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 21 0^{\circ}, x = 33 0^{\circ}, x = 0.84806 \dots, x = 18 0^{\circ} - 0.84806 \dots

Solution steps

4 cos (2 x) + 2 sin (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

Subtract

1

from both sides

4 cos (2 x) + 2 sin (x) - 1 = 0

Rewrite using trig identities

3 + 2 sin (x) - 8 sin^{2} (x) = 0

Solve by substitution

sin (x) = - \frac{1}{2}, sin (x) = \frac{3}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 36 0^{\circ} : x = 21 0^{\circ}, x = 33 0^{\circ}

sin (x) = \frac{3}{4}, 0 \leq x \leq 36 0^{\circ} : x = arcsin (\frac{3}{4}), x = 18 0^{\circ} - arcsin (\frac{3}{4})

Combine all the solutions

x = 21 0^{\circ}, x = 33 0^{\circ}, x = arcsin (\frac{3}{4}), x = 18 0^{\circ} - arcsin (\frac{3}{4})

Show solutions in decimal form

x = 21 0^{\circ}, x = 33 0^{\circ}, x = 0.84806 \dots, x = 18 0^{\circ} - 0.84806 \dots

3 csc (θ) + 5 = 0

cos (\frac{x}{3}) - cos (\frac{x}{15}) = 0

(2 cos (x) - 3) (2 sin (x) + 1) = 0

y, x = π - arcsin (y)

arctan (x) = 3

What is the general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 ?
The general solution for 4cos(2x)+2sin(x)=1,0<= x<= 360 is x=210,x=330,x=0.84806…,x=180-0.84806…