What is the general solution for -8sin(x)+8cos(2x)=0 ?

The general solution for -8sin(x)+8cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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-8sin(x)+8cos(2x)=0

- 8 sin (x) + 8 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

- 8 sin (x) + 8 cos (2 x) = 0

Rewrite using trig identities

(1 - 2 sin^{2} (x)) \cdot 8 - 8 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

tan^{2} (x) + 2 sec (x) + 2 = 0

sin (θ) = 1 + cos (θ)

7 cos (2 x) = 7 cos^{2} (x) - 3

- cos^{2} (θ) + 1 = 0

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

What is the general solution for -8sin(x)+8cos(2x)=0 ?
The general solution for -8sin(x)+8cos(2x)=0 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin