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

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

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5sin(2x)=5cos(x),0<= x<= 2pi

5 sin (2 x) = 5 cos (x), 0 \leq x \leq 2 π

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

Solution steps

5 sin (2 x) = 5 cos (x), 0 \leq x \leq 2 π

Subtract

5 cos (x)

from both sides

5 sin (2 x) - 5 cos (x) = 0

Rewrite using trig identities

- 5 cos (x) + 10 cos (x) sin (x) = 0

Factor

- 5 cos (x) + 10 cos (x) sin (x) : 5 cos (x) (2 sin (x) - 1)

5 cos (x) (2 sin (x) - 1) = 0

Solving each part separately

cos (x) = 0 or 2 sin (x) - 1 = 0

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

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

Combine all the solutions

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

x, f = cos (x) cos (h y)

tan (x) = - \frac{1}{10}

tan (4 x) \cdot cot (x + 60) = 1

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

2 - 2 cos (x) = - 6 cos (x)

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