What is the general solution for 8sin^2(x)2cos(x)=7,0<= x<= 2pi ?

The general solution for 8sin^2(x)2cos(x)=7,0<= x<= 2pi is No Solution for x\in\mathbb{R}

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8sin^2(x)2cos(x)=7,0<= x<= 2pi

8 sin^{2} (x) 2 cos (x) = 7, 0 \leq x \leq 2 π

No Solution for x \in R

Solution steps

8 sin^{2} (x) \cdot 2 cos (x) = 7, 0 \leq x \leq 2 π

Subtract

7

from both sides

16 sin^{2} (x) cos (x) - 7 = 0

Rewrite using trig identities

- 7 + (1 - cos^{2} (x)) \cdot 16 cos (x) = 0

Solve by substitution

cos (x) \approx - 1.17189 \dots

cos (x) = - 1.17189 \dots, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

No Solution for x \in R

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

9 cos (2 x) = 9 cos^{2} (x) - 4

2 - cos (x) = 0

sin (\frac{1}{x}) = 1

2 sin (θ) - cos (θ) = 0

What is the general solution for 8sin^2(x)2cos(x)=7,0<= x<= 2pi ?
The general solution for 8sin^2(x)2cos(x)=7,0<= x<= 2pi is No Solution for x\in\mathbb{R}