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

The general solution for 2sin(x)cos(x)=cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=0.46364…+pin

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2sin(x)cos(x)=cos^2(x)

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.46364 \dots + πn

Solution steps

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

Subtract

cos^{2} (x)

from both sides

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

Factor

2 sin (x) cos (x) - cos^{2} (x) : cos (x) (2 sin (x) - cos (x))

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

Solving each part separately

cos (x) = 0 or 2 sin (x) - cos (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

2 sin (x) - cos (x) = 0 : x = arctan (\frac{1}{2}) + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arctan (\frac{1}{2}) + πn

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.46364 \dots + πn

- 1 + sin^{2} (a) + cos^{2} (a) = cos^{2} (a)

\frac{sin ( x )}{1 + cos ( x )} = 3 cot (\frac{x}{2})

sin (θ) = \frac{3}{2}, 0 \leq θ \leq 4 π

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

3 sin^{2} (x) + 8 cos^{2} (x) = 6

What is the general solution for 2sin(x)cos(x)=cos^2(x) ?
The general solution for 2sin(x)cos(x)=cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=0.46364…+pin