Question

Solve the equation $\sqrt{6 - cos x + 7 sin^{2} x} + cos x = 0$.

Answer 1

Given that

$\sqrt{6-Cosx+7Sin^{2}x}+Cosx=0$

$\sqrt{6-Cosx+7Sin^{2}x}=-Cosx$

Squaring both sides, we get

6 - Cos x + 7 Sin²x = Cos²x

6 - Cos x + 7 (1 - Cos²x) = Cos²x

6 - Cos x + 7  - 7 Cos²x = Cos²x

8 Cos²x + Cos x - 13 = 0

Using Quadratic Formula, we get

$Cosx=\frac{-1\pm\sqrt{1-4(8)(-13)} }{2(8)} \\\\Cosx=\frac{-1\pm\sqrt{1-4(8)(-13)} }{16}\\\\Cosx=\frac{-1\pm20.4 }{16}\\\\Cosx=\frac{-1\pm20.4 }{16}AND\\\\Cosx=\frac{-1\pm20.4 }{16}$

Cos x = 1.2125 And Cos x = - 1.26

In both cases, its Solution is not Possible.

Answer 2

After transposing Cos x to right side,

Squaring both sides, we get

6 - Cos x + 7 Sin²x = Cos²x

6 - Cos x + 7 (1 - Cos²x) = Cos²x

6 - Cos x + 7  - 7 Cos²x = Cos²x

8 Cos²x + Cos x - 13 = 0

Using Quadratic Formula, we get

Cos x = 1.2125 And Cos x = - 1.26

In both cases, its Solution is not Possible.: