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

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

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tan^2(x)-4sin(x)+4=0

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

No Solution for x \in R

Solution steps

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

Add

4 sin (x)

to both sides

tan^{2} (x) + 4 = 4 sin (x)

Square both sides

(tan^{2} (x) + 4)^{2} = (4 sin (x))^{2}

Subtract

(4 sin (x))^{2}

from both sides

(tan^{2} (x) + 4)^{2} - 16 sin^{2} (x) = 0

Rewrite using trig identities

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 sin^{2} (x) = 0

Factor

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )})^{2} - 16 sin^{2} (x) : (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x)) (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x))

(4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x)) (4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x)) = 0

Solving each part separately

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x) = 0 or 4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x) = 0

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} + 4 sin (x) = 0 :

No Solution

4 + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - 4 sin (x) = 0 :

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

No Solution for x \in R

\frac{1}{sin ( x )} + \frac{1}{cos ( x )} = 0

1 - cos (2 x) = 0

sin (x) = \frac{37}{64}

1 = 4 tan (- \frac{4}{81} + c)

15 cos^{2} (x) + 7 cos (x) - 2 = 0

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