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

The general solution for 4cos^2(x)+4cos(x)=1,0<= x<360 is x=1.36217…,x=360-1.36217…

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

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

x = 1.36217 \dots, x = 36 0^{\circ} - 1.36217 \dots

Solution steps

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

Solve by substitution

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

cos (x) = \frac{- 1 + 2}{2}, 0 \leq x < 36 0^{\circ} : x = arccos (\frac{2 - 1}{2}), x = 36 0^{\circ} - arccos (\frac{2 - 1}{2})

cos (x) = - \frac{1 + 2}{2}, 0 \leq x < 36 0^{\circ} :

No Solution

Combine all the solutions

x = arccos (\frac{2 - 1}{2}), x = 36 0^{\circ} - arccos (\frac{2 - 1}{2})

Show solutions in decimal form

x = 1.36217 \dots, x = 36 0^{\circ} - 1.36217 \dots

8 cos^{2} (x) = 4

4 cos (3 x) = 0

sin^{2} (x) = sin^{2} (\frac{x}{2})

cos (x) = \frac{6}{8}

10 sin (a) + 16 = 3 sin (a) + 9

What is the general solution for 4cos^2(x)+4cos(x)=1,0<= x<360 ?
The general solution for 4cos^2(x)+4cos(x)=1,0<= x<360 is x=1.36217…,x=360-1.36217…