What is the general solution for solvefor x,f=2cos(3x^2-1)entoncesf ?

The general solution for solvefor x,f=2cos(3x^2-1)entoncesf is

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solvefor x,f=2cos(3x^2-1)entoncesf

solve for

x, f = 2 cos (3 x^{2} - 1) e n t o n ces f

x = \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

Solution steps

f = 2 cos (3 x^{2} - 1) e n t o n ces f

Switch sides

2 cos (3 x^{2} - 1) e n t o n ces f = f

Divide both sides by

2 e n t o n ces f; c \neq = 0

cos (3 x^{2} - 1) = \frac{1}{2 e ^{2} c t os n ^{2}}; c \neq = 0

Apply trig inverse properties

3 x^{2} - 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk, 3 x^{2} - 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk

Solve

3 x^{2} - 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk : x = \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

Solve

3 x^{2} - 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk : x = \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

x = \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

tan (a) = \frac{5}{8}

x, z = tan (\frac{x}{2})

tan (\frac{x}{2}) = 3, 0 \leq x \leq 2 π

0 = 4 sin (2 θ)

tan (θ) + 3 = 0

What is the general solution for solvefor x,f=2cos(3x^2-1)entoncesf ?
The general solution for solvefor x,f=2cos(3x^2-1)entoncesf is