What is the general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi ?

The general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi is x=pi+0.46619…,x=-0.46619…+2pi

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-4sin(x)=cos^2(x)+1,0<= x<= 2pi

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

x = π + 0.46619 \dots, x = - 0.46619 \dots + 2 π

Solution steps

- 4 sin (x) = cos^{2} (x) + 1, 0 \leq x \leq 2 π

Subtract

cos^{2} (x) + 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = 2 + 6, sin (x) = 2 - 6

sin (x) = 2 + 6, 0 \leq x \leq 2 π :

No Solution

sin (x) = 2 - 6, 0 \leq x \leq 2 π : x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

Combine all the solutions

x = π + arcsin (6 - 2), x = arcsin (2 - 6) + 2 π

Show solutions in decimal form

x = π + 0.46619 \dots, x = - 0.46619 \dots + 2 π

3 sin (x) - 3 cos (x) = 3

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

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

sin (x) = \frac{1.2}{10}

t, x = - 3 cos (π t)

What is the general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi ?
The general solution for -4sin(x)=cos^2(x)+1,0<= x<= 2pi is x=pi+0.46619…,x=-0.46619…+2pi