What is the general solution for 2cos^2(x)+sin(x)=0 ?

The general solution for 2cos^2(x)+sin(x)=0 is x=-0.89590…+2pin,x=pi+0.89590…+2pin

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2cos^2(x)+sin(x)=0

2 cos^{2} (x) + sin (x) = 0

x = - 0.89590 \dots + 2 πn, x = π + 0.89590 \dots + 2 πn

Solution steps

2 cos^{2} (x) + sin (x) = 0

Rewrite using trig identities

sin (x) + (1 - sin^{2} (x)) \cdot 2 = 0

Solve by substitution

sin (x) = - \frac{- 1 + 17}{4}, sin (x) = \frac{1 + 17}{4}

sin (x) = - \frac{- 1 + 17}{4} : x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{1 + 17}{4} :

No Solution

Combine all the solutions

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

Show solutions in decimal form

x = - 0.89590 \dots + 2 πn, x = π + 0.89590 \dots + 2 πn

2 cos^{2} (x) + sin (x) = 5

x, Y = 0.5 sin (3.07 x - 2.4 t + 0.59)

\frac{( 1 + tan ( x ) )}{sec ( x )} = 2, x

sin (x - 3 0^{\circ}) cos (x - 3 0^{\circ}) = \frac{3}{4}

cos (θ) = - \frac{7}{15}, cos (\frac{θ}{2}), 18 0^{\circ} < θ < 27 0^{\circ}

What is the general solution for 2cos^2(x)+sin(x)=0 ?
The general solution for 2cos^2(x)+sin(x)=0 is x=-0.89590…+2pin,x=pi+0.89590…+2pin