What is the general solution for 1+sin(x)=2cos^2(x),0<= x<5 ?

The general solution for 1+sin(x)=2cos^2(x),0<= x<5 is x= pi/6 ,x=(5pi)/6 ,x=(3pi)/2

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1+sin(x)=2cos^2(x),0<= x<5

1 + sin (x) = 2 cos^{2} (x), 0 \leq x < 5

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

Solution steps

1 + sin (x) = 2 cos^{2} (x), 0 \leq x < 5

Subtract

2 cos^{2} (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = \frac{1}{2}, 0 \leq x < 5 : x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = - 1, 0 \leq x < 5 : x = \frac{3 π}{2}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{3 π}{2}

2 - 3 cos (x) = 5 + 3 cos (x)

sinh (x) = \frac{6}{5}

tan (2 x) tan (x) = 1

\frac{csc ^{2} ( x )}{4} = 4 sin^{2} (x)

6 sec^{2} (x) - 3 cos (x) - 10 = sec (x)

What is the general solution for 1+sin(x)=2cos^2(x),0<= x<5 ?
The general solution for 1+sin(x)=2cos^2(x),0<= x<5 is x= pi/6 ,x=(5pi)/6 ,x=(3pi)/2