solve the first equation in each pair of equations or y and /or z. Then use the same strategy to solve the second equation for x in the interval 0<= x<= 2pie.
y^2= 1/3, tan^2x = 1/3
yz= y, tan x sec x = tanx
y^2 = 1/3
y = ± 1/√3
so tan^2 x = 1/3
tanx = ± 1/√3
so x could be in all 4 quadrants
knowing the ratio of the 30-60-90 triangle is 1:√3:2
x must be 30º or pi/6 radians
so
x = pi/6 in I
x = pi - pi/6 = 5pi/6 in II
x = pi + pi/6 = 7pi/6 in III
x = 2pi - pi/6 = 11pi/6 in IV
for tan x sec x = tanx
divide by tanx
secx = 1
cosx = 1
Where is the cosine graph 1 ?
I will let you finish it.
To solve the first equation, we have y^2 = 1/3. Taking the square root of both sides, we get y = ±√(1/3). Since the square root of a positive number has two possible values (positive and negative), we have two solutions: y = √(1/3) and y = -√(1/3).
To solve the second equation, tan^2x = 1/3, we first take the square root of both sides, which gives us tanx = ±√(1/3). However, we need to find the value of x in the interval 0 <= x <= 2π (or 0 <= x <= 360 degrees).
Using a calculator or a trigonometric table, we can find the values of x whose tangent is ±√(1/3). One solution is x = 30 degrees or π/6 radians, since tan(30 degrees) = tan(π/6 radians) = √(1/3). Another solution is x = 150 degrees or 5π/6 radians, since tan(150 degrees) = tan(5π/6 radians) = -√(1/3).
For the second pair of equations, we have yz = y. We can divide both sides by y, which gives us z = 1. So, the solution for this equation is z = 1.
For the equation tan x sec x = tan x, we can divide both sides by tan x, which gives us sec x = 1. The cosine function is the reciprocal of the secant function, so sec x = 1 is equivalent to cos x = 1/1 = 1.
The only angle in the interval 0 <= x <= 2π (or 0 <= x <= 360 degrees) where the cosine value is 1 is x = 0 degrees or 0 radians.
Therefore, the solutions for the second equation are z = 1 and x = 0.