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

The general solution for 2sin^2(x/2)-cos(x/2)=0 is x=2*0.67488…+4pin,x=4pi-2*0.67488…+4pin

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2sin^2(x/2)-cos(x/2)=0

2 sin^{2} (\frac{x}{2}) - cos (\frac{x}{2}) = 0

x = 2 \cdot 0.67488 \dots + 4 πn, x = 4 π - 2 \cdot 0.67488 \dots + 4 πn

Solution steps

2 sin^{2} (\frac{x}{2}) - cos (\frac{x}{2}) = 0

Rewrite using trig identities

- cos (\frac{x}{2}) + (1 - cos^{2} (\frac{x}{2})) \cdot 2 = 0

Solve by substitution

cos (\frac{x}{2}) = - \frac{1 + 17}{4}, cos (\frac{x}{2}) = \frac{17 - 1}{4}

cos (\frac{x}{2}) = - \frac{1 + 17}{4} :

No Solution

cos (\frac{x}{2}) = \frac{17 - 1}{4} : x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

Combine all the solutions

x = 2 arccos (\frac{17 - 1}{4}) + 4 πn, x = 4 π - 2 arccos (\frac{17 - 1}{4}) + 4 πn

Show solutions in decimal form

x = 2 \cdot 0.67488 \dots + 4 πn, x = 4 π - 2 \cdot 0.67488 \dots + 4 πn

cos (\frac{π}{4} - x) = \frac{12}{13}

x, y = f cos (x)

3 cot (2 x) = - 1, 0 \leq x \leq π

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

(cot (θ) - 1) (2 sin (θ) + 3) = 0

What is the general solution for 2sin^2(x/2)-cos(x/2)=0 ?
The general solution for 2sin^2(x/2)-cos(x/2)=0 is x=2*0.67488…+4pin,x=4pi-2*0.67488…+4pin