4cos(x^2)-1=0

Solution

4 cos (x^{2}) - 1 = 0

x = 1.31811 \dots + 2 πn, x = - 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = - 2 π - 1.31811 \dots + 2 πn

Solution steps

4 cos (x^{2}) - 1 = 0

Move

1

to the right side

4 cos (x^{2}) = 1

Divide both sides by

4

cos (x^{2}) = \frac{1}{4}

Apply trig inverse properties

x^{2} = arccos (\frac{1}{4}) + 2 πn, x^{2} = 2 π - arccos (\frac{1}{4}) + 2 πn

Solve

x^{2} = arccos (\frac{1}{4}) + 2 πn : x = arccos (\frac{1}{4}) + 2 πn, x = - arccos (\frac{1}{4}) + 2 πn

Solve

x^{2} = 2 π - arccos (\frac{1}{4}) + 2 πn : x = 2 π - arccos (\frac{1}{4}) + 2 πn, x = - 2 π - arccos (\frac{1}{4}) + 2 πn

x = arccos (\frac{1}{4}) + 2 πn, x = - arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn, x = - 2 π - arccos (\frac{1}{4}) + 2 πn

Show solutions in decimal form

x = 1.31811 \dots + 2 πn, x = - 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = - 2 π - 1.31811 \dots + 2 πn

2 cos^{2} (x) - 7 cos (x) = - 3, 0 \leq x \leq 2 π

sin (2 x) = \frac{( 4 m - 9 )}{8}

sin (2 x) = - 2

sin (x) = sin (9 0^{\circ} - x)

2 sin (2 θ) - 3 - \frac{6}{sin ( 2 θ ) - 1} = 0