What is the general solution for sqrt(3)sin(x)-((sqrt(3)sin(x))-1)=1n ?

The general solution for sqrt(3)sin(x)-((sqrt(3)sin(x))-1)=1n is No Solution for x\in\mathbb{R}

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sqrt(3)sin(x)-((sqrt(3)sin(x))-1)=1n

3 sin (x) - ((3 sin (x)) - 1) = 1 n

No Solution for x \in R

Solution steps

3 sin (x) - ((3 sin (x)) - 1) = 1 \cdot n

Solve by substitution

3 sin (x) - (3 sin (x) - 1) = 1 \cdot n

Let:

sin (x) = u

3 u - (3 u - 1) = 1 \cdot n

3 u - (3 u - 1) = 1 \cdot n :

True for all

u; 1 = n

3 u - (3 u - 1) = 1 \cdot n

Multiply:

1 \cdot n = n

3 u - (3 u - 1) = n

Expand

3 u - (3 u - 1) : 1

3 u - (3 u - 1)

- (3 u - 1) : - 3 u + 1

- (3 u - 1)

Distribute parentheses

= - (3 u) - (- 1)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 3 u + 1

= 3 u - 3 u + 1

Add similar elements:

3 u - 3 u = 0

= 1

1 = n

Both sides are equal

True for all u; 1 = n

Substitute back

u = sin (x)

True for all sin (x); 1 = n

True for all sin (x); 1 = n

sin (x) =

True for all

u : x = arcsin (True for all u) + 2 πk, x = π + arcsin (- True for all u) + 2 πk

sin (x) = True for all u

Apply trig inverse properties

sin (x) = True for all u

General solutions for

sin (x) =

True for all

u

sin (x) = a \Rightarrow x = arcsin (a) + 2 πk, x = π + arcsin (a) + 2 πk

x = arcsin (True for all u) + 2 πk, x = π + arcsin (- True for all u) + 2 πk

x = arcsin (True for all u) + 2 πk, x = π + arcsin (- True for all u) + 2 πk

Combine all the solutions

x = arcsin (True for all u) + 2 πk, x = π + arcsin (- True for all u) + 2 πk

Since the equation is undefined for:

arcsin (True for all u) + 2 πk, π + arcsin (- True for all u) + 2 πk

No Solution for x \in R

What is the general solution for sqrt(3)sin(x)-((sqrt(3)sin(x))-1)=1n ?
The general solution for sqrt(3)sin(x)-((sqrt(3)sin(x))-1)=1n is No Solution for x\in\mathbb{R}