What is the general solution for 2sin^2(x)+cos(x)-1=0,interval^0,2pi ?

The general solution for 2sin^2(x)+cos(x)-1=0,interval^0,2pi is No Solution for x\in\mathbb{R}

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2sin^2(x)+cos(x)-1=0,interval^0,2pi

2 sin^{2} (x) + cos (x) - 1 = 0, in t er v a l^{0}, 2 π

No Solution for x \in R

Solution steps

2 sin^{2} (x) + cos (x) - 1 = 0, in t er v a l^{0}, 2 π

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - \frac{1}{2}, in t er v a l^{0} :

No Solution

cos (x) = 1, in t er v a l^{0} :

No Solution

Combine all the solutions

No Solution for x \in R

tan (x) = \frac{100}{23}

4 sin (\frac{x}{2}) + 4 cos (x) = 0

cos (θ) cos (θ) = cot (θ)

3 sin (θ) - cos (2 θ) = 1

sin (θ) = \frac{28}{53}, sin (2 θ)

What is the general solution for 2sin^2(x)+cos(x)-1=0,interval^0,2pi ?
The general solution for 2sin^2(x)+cos(x)-1=0,interval^0,2pi is No Solution for x\in\mathbb{R}