What is the general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 ?

The general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 is x=2.53724…+2pin,x=-2.53724…+2pin

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2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2

2 + cos (x) = 3 \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

x = 2.53724 \dots + 2 πn, x = - 2.53724 \dots + 2 πn

Solution steps

2 + cos (x) = 3 \cdot \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

Subtract

3 \frac{cos ^{2} ( x )}{2} + \frac{sin ^{2} ( x )}{2}

from both sides

2 + cos (x) - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2} = 0

Simplify

2 + cos (x) - \frac{3 cos ^{2} ( x ) + sin ^{2} ( x )}{2} : \frac{4 + 2 cos ( x ) - 3 cos ^{2} ( x ) - sin ^{2} ( x )}{2}

\frac{4 + 2 cos ( x ) - 3 cos ^{2} ( x ) - sin ^{2} ( x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 + 2 cos (x) - 3 cos^{2} (x) - sin^{2} (x) = 0

Rewrite using trig identities

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

Solve by substitution

cos (x) = - \frac{- 1 + 7}{2}, cos (x) = \frac{1 + 7}{2}

cos (x) = - \frac{- 1 + 7}{2} : x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

cos (x) = \frac{1 + 7}{2} :

No Solution

Combine all the solutions

x = arccos (- \frac{- 1 + 7}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 7}{2}) + 2 πn

Show solutions in decimal form

x = 2.53724 \dots + 2 πn, x = - 2.53724 \dots + 2 πn

sin^{2} (x) + sin^{6} (x) = 3 cos^{2} (2 x)

cot^{2} (x) - 7 cot (x) + 10 = 0

x, r - 2 s + t = sin (2 x + 3 y)

sin^{5} (a) = 16 sin^{5} (a) - 20 sin^{3} (a) + 5 sin (a)

tan (b) = \frac{1}{2}

What is the general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 ?
The general solution for 2+cos(x)=3(cos^2(x))/2+(sin^2(x))/2 is x=2.53724…+2pin,x=-2.53724…+2pin