What is the general solution for 1/2 cos^2(2x)+tan(162)=0 ?

The general solution for 1/2 cos^2(2x)+tan(162)=0 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/2 cos^2(2x)+tan(162)=0

\frac{1}{2} cos^{2} (2 x) + tan (16 2^{\circ}) = 0

x = \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = \frac{2.50837 \dots}{2} + 18 0^{\circ} n, x = - \frac{2.50837 \dots}{2} + 18 0^{\circ} n

Solution steps

\frac{1}{2} cos^{2} (2 x) + tan (16 2^{\circ}) = 0

tan (16 2^{\circ}) = - \frac{5 - 2 5}{5}

\frac{1}{2} cos^{2} (2 x) + - \frac{5 - 2 5}{5} = 0

Rewrite using trig identities

\frac{1}{2 sec ^{2} ( 2 x )} - \frac{- 2 + 5}{5} = 0

Solve by substitution

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2}, sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2}

sec (2 x) = \frac{( 2 + 5 ) 5 - 2 5}{2} : x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

sec (2 x) = - \frac{( 2 + 5 ) 5 - 2 5}{2} : x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

Combine all the solutions

x = \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{arcsec ( \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n, x = - \frac{arcsec ( - \frac{( 2 + 5 ) 5 - 2 5}{2} )}{2} + 18 0^{\circ} n

Show solutions in decimal form

x = \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = 18 0^{\circ} - \frac{0.63322 \dots}{2} + 18 0^{\circ} n, x = \frac{2.50837 \dots}{2} + 18 0^{\circ} n, x = - \frac{2.50837 \dots}{2} + 18 0^{\circ} n

1 + cot^{2} (a) = tan^{2} (a)

tan (α) = \frac{12}{6 3}

1 = cos^{2} (α) - 2 cos (α) sin (a) + sin^{2} (α)

4 sin (θ) = 1

sin (x + \frac{π}{2}) = 0.6

What is the general solution for 1/2 cos^2(2x)+tan(162)=0 ?
The general solution for 1/2 cos^2(2x)+tan(162)=0 is