What is the general solution for 1+sin(x)=2cos(x) ?

The general solution for 1+sin(x)=2cos(x) is x=0.64350…+2pin,x=(3pi)/2+2pin

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1+sin(x)=2cos(x)

1 + sin (x) = 2 cos (x)

x = 0.64350 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

1 + sin (x) = 2 cos (x)

Square both sides

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

Subtract

(2 cos (x))^{2}

from both sides

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

Rewrite using trig identities

- 3 + 2 sin (x) + 5 sin^{2} (x) = 0

Solve by substitution

sin (x) = \frac{3}{5}, sin (x) = - 1

sin (x) = \frac{3}{5} : x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn, x = \frac{3 π}{2} + 2 πn

Verify solutions by plugging them into the original equation

x = arcsin (\frac{3}{5}) + 2 πn, x = \frac{3 π}{2} + 2 πn

Show solutions in decimal form

x = 0.64350 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

3 sin (x) = 1 + cos (x)

sin (\frac{1}{2} x) = 0

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

\frac{sin ( x )}{cos ( x )} = 1

9 (1 - cos (θ)) = sin^{2} (θ)

What is the general solution for 1+sin(x)=2cos(x) ?
The general solution for 1+sin(x)=2cos(x) is x=0.64350…+2pin,x=(3pi)/2+2pin