What is the general solution for -14sin(x)+14cos(2x)=0 ?

The general solution for -14sin(x)+14cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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-14sin(x)+14cos(2x)=0

- 14 sin (x) + 14 cos (2 x) = 0

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

Solution steps

- 14 sin (x) + 14 cos (2 x) = 0

Rewrite using trig identities

(1 - 2 sin^{2} (x)) \cdot 14 - 14 sin (x) = 0

Solve by substitution

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

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

2 cos (2 x) + 2 = 0, 0 \leq x \leq 2 π

sin (x) = \frac{10}{25}

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

4 tan^{2} (x) + 7 tan (x) - 4 = 0

tan (θ) = - 1 - 3

What is the general solution for -14sin(x)+14cos(2x)=0 ?
The general solution for -14sin(x)+14cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin