What is the general solution for 4sin^2(x)-3=4,sin(2x)-3=0 ?

The general solution for 4sin^2(x)-3=4,sin(2x)-3=0 is No Solution for x\in\mathbb{R}

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Popular >

4sin^2(x)-3=4,sin(2x)-3=0

Solution

4 sin^{2} (x) - 3 = 4, sin (2 x) - 3 = 0

Solution

No Solution for x \in R

Solution steps

4 sin^{2} (x) - 3 = 4, sin (2 x) - 3 = 0

Solve by substitution

4 sin^{2} (x) - 3 = 4

Let:

sin (x) = u

4 u^{2} - 3 = 4

4 u^{2} - 3 = 4 : u = \frac{7}{2}, u = - \frac{7}{2}

4 u^{2} - 3 = 4

Move

3

to the right side

4 u^{2} - 3 = 4

Add

3

to both sides

4 u^{2} - 3 + 3 = 4 + 3

Simplify

4 u^{2} = 7

4 u^{2} = 7

Divide both sides by

4

4 u^{2} = 7

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{7}{4}

Simplify

u^{2} = \frac{7}{4}

u^{2} = \frac{7}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{7}{4}, u = - \frac{7}{4}

\frac{7}{4} = \frac{7}{2}

\frac{7}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{7}{2}

- \frac{7}{4} = - \frac{7}{2}

- \frac{7}{4}

Simplify

\frac{7}{4} : \frac{7}{2}

\frac{7}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{7}{2}

= - \frac{7}{2}

u = \frac{7}{2}, u = - \frac{7}{2}

Substitute back

u = sin (x)

sin (x) = \frac{7}{2}, sin (x) = - \frac{7}{2}

sin (x) = \frac{7}{2}, sin (x) = - \frac{7}{2}

sin (x) = \frac{7}{2}, sin (2 x) - 3 = 0 :

No Solution

sin (x) = \frac{7}{2}, sin (2 x) - 3 = 0

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = - \frac{7}{2}, sin (2 x) - 3 = 0 :

No Solution

sin (x) = - \frac{7}{2}, sin (2 x) - 3 = 0

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(x)-3=4,sin(2x)-3=0 ?
The general solution for 4sin^2(x)-3=4,sin(2x)-3=0 is No Solution for x\in\mathbb{R}