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- Equation:
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Equation а²+6а+9+(а+3)(3-а)
- Equation (-2cos²x+sinx+1)×log0,5(-0,8cosx)=0
- Equation sin^3x-sin^2x+sinx-1=0
- Equation 3x-3x2=-26x=10
- Express {x} in terms of y where:
- 5*x-5*y=20
- 8*x-7*y=-13
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Identical expressions
- (-2cos²x+sinx+ one)×log zero , five (- zero ,8cosx)=0
- ( minus 2 co sinus of e of ²x plus sinus of x plus 1)× logarithm of 0,5( minus 0,8 co sinus of e of x) equally 0
- ( minus 2 co sinus of e of ²x plus sinus of x plus one)× logarithm of zero , five ( minus zero ,8 co sinus of e of x) equally 0
- -2cos²x+sinx+1×log0,5-0,8cosx=0
- (-2cos²x+sinx+1)×log0,5(-0,8cosx)=O
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Similar expressions
- (2cos²x+sinx+1)×log0,5(-0,8cosx)=0
- (-2cos²x-sinx+1)×log0,5(-0,8cosx)=0
- (-2cos²x+sinx-1)×log0,5(-0,8cosx)=0
- (-2cos²x+sinx+1)×log0,5(0,8cosx)=0
(-2cos²x+sinx+1)×log0,5(-0,8cosx)=0 equation
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The solution
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/ 2 \ -4*cos(x)
\- 2*cos (x) + sin(x) + 1/*log(0.0)*5*--------- = 0
5
$$5 \left(\left(\sin{\left(x \right)} - 2 \cos^{2}{\left(x \right)}\right) + 1\right) \log{\left(0.0 \right)} \left(- \frac{4 \cos{\left(x \right)}}{5}\right) = 0$$
Detail solution
Given the equation
$$5 \left(\left(\sin{\left(x \right)} - 2 \cos^{2}{\left(x \right)}\right) + 1\right) \log{\left(0 \right)} \left(- \frac{4 \cos{\left(x \right)}}{5}\right) = 0$$
transform
$$\tilde{\infty} \left(\sin^{2}{\left(x \right)} + \sin{\left(x \right)} + 1\right) \cos{\left(x \right)} = 0$$
$$- 4 \left(2 \sin^{2}{\left(x \right)} + \sin{\left(x \right)} - 1\right) \log{\left(0 \right)} \cos{\left(x \right)} = 0$$
Do replacement
$$w = \sin{\left(x \right)}$$
Expand the expression in the equation
$$\tilde{\infty} \left(2 w^{2} + w - 1\right) \cos{\left(x \right)} = 0$$
We get the quadratic equation
$$\tilde{\infty} w^{2} \cos{\left(x \right)} + \tilde{\infty} w \cos{\left(x \right)} + \tilde{\infty} \cos{\left(x \right)} = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \tilde{\infty} \cos{\left(x \right)}$$
$$b = \tilde{\infty} \cos{\left(x \right)}$$
$$c = \tilde{\infty} \cos{\left(x \right)}$$
, then
Because D = 0, then the equation has one root.
do backward replacement
$$\sin{\left(x \right)} = w$$
Given the equation
$$\sin{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
Or
$$x = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
, where n - is a integer
substitute w:
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(w_{1} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(\text{NaN} \right)}$$
$$x_{2} = 2 \pi n - \operatorname{asin}{\left(w_{1} \right)} + \pi$$
$$x_{2} = 2 \pi n + \pi - \operatorname{asin}{\left(\text{NaN} \right)}$$
$$5 \left(\left(\sin{\left(x \right)} - 2 \cos^{2}{\left(x \right)}\right) + 1\right) \log{\left(0 \right)} \left(- \frac{4 \cos{\left(x \right)}}{5}\right) = 0$$
transform
$$\tilde{\infty} \left(\sin^{2}{\left(x \right)} + \sin{\left(x \right)} + 1\right) \cos{\left(x \right)} = 0$$
$$- 4 \left(2 \sin^{2}{\left(x \right)} + \sin{\left(x \right)} - 1\right) \log{\left(0 \right)} \cos{\left(x \right)} = 0$$
Do replacement
$$w = \sin{\left(x \right)}$$
Expand the expression in the equation
$$\tilde{\infty} \left(2 w^{2} + w - 1\right) \cos{\left(x \right)} = 0$$
We get the quadratic equation
$$\tilde{\infty} w^{2} \cos{\left(x \right)} + \tilde{\infty} w \cos{\left(x \right)} + \tilde{\infty} \cos{\left(x \right)} = 0$$
This equation is of the form
a*w^2 + b*w + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = \tilde{\infty} \cos{\left(x \right)}$$
$$b = \tilde{\infty} \cos{\left(x \right)}$$
$$c = \tilde{\infty} \cos{\left(x \right)}$$
, then
D = b^2 - 4 * a * c =
(±oo*cos(x))^2 - 4 * (±oo*cos(x)) * (±oo*cos(x)) = 0
Because D = 0, then the equation has one root.
w = -b/2a = -±oo*cos(x)/2/(±oo*cos(x))
False
do backward replacement
$$\sin{\left(x \right)} = w$$
Given the equation
$$\sin{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
Or
$$x = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
, where n - is a integer
substitute w:
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(w_{1} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(\text{NaN} \right)}$$
False
$$x_{2} = 2 \pi n - \operatorname{asin}{\left(w_{1} \right)} + \pi$$
$$x_{2} = 2 \pi n + \pi - \operatorname{asin}{\left(\text{NaN} \right)}$$
False